1 calculate the boundaries. Another miracle boundary. Algorithm for calculating limits

Sound another miracle limit write down in this form:

\begin(equation) \lim_(x\to\infty)\left(1+\frac(1)(x)\right)^x=e\end(equation)

The number $e$, shown at the right side of equality (1), is irrational. An approximation of the value of this number is: $ e \ approx (2 (,) 718281828459045) $. If I replace $t=\frac(1)(x)$, then formula (1) can be rewritten as follows:

\begin(equation) \lim_(t\to(0))\biggl(1+t\biggr)^(\frac(1)(t))=e\end(equation)

As for the first miraculous boundary, it does not matter which one should replace the change $x$ in formula (1) or replace the change $t$ in formula (2). Golovne is a vikonnannya of two minds:

1. Substava step (tobto viraz at the arches of formulas (1) and (2)) may be singled out;
2. The step indicator (either $x$ for formula (1) or $\frac(1)(t)$ for formula (2)) may not incur inconsistency.

To say that another miraculous boundary line is the insignificance of $1^\infty$. Please note that formula (1) is not specified, about the very inconsistency ($+\infty$ or $-\infty$) can be found. In skin disorders, the formula (1) is true. For formula (2), $t$ can be changed to zero as evil, i on the right.

I’ll tell you that it’s also a little brown heritage from another miraculous boundary. Apply another wand of another miraculous boundary, like a legacy from the new one, more popular with the styling of standard typical roses and control robots.

Butt #1

Calculate between $\lim_(x\to\infty)\left(\frac(3x+1)(3x-5)\right)^(4x+7)$.

It is immediately significant that the basis of the step (that is, $\frac(3x+1)(3x-5)$) is not the same:

$$ \lim_(x\to\infty)\frac(3x+1)(3x-5)=\left|\frac(\infty)(\infty)\right| =\lim_(x\to\infty)\frac(3+\frac(1)(x))(3-\frac(5)(x)) =\frac(3+0)(3-0) = 1. $$

Whom has a step (viraz $4x+7$) of perfect inconsistency, tobto. $\lim_(x\to\infty)(4x+7)=\infty$.

The step of the pragna is alone, the indicator of the step is to the point of inconsistency, tobto. We can rightly see $1^\infty$ being insignificant. Zastosuєmo formula for razkrittya cієї neznachennostі. In the basis of the step of the formula, the following is $1+\frac(1)(x)$, and in the example we have suggested, the step of the step is as follows: $\frac(3x+1)(3x-5)$. To that first task, I formally assign the virase $\frac(3x+1)(3x-5)$ to look at $1+\frac(1)(x)$. For the cob dodamo and see one:

$$ \lim_(x\to\infty)\left(\frac(3x+1)(3x-5)\right)^(4x+7) =|1^\infty| =\lim_(x\to\infty)\left(1+\frac(3x+1)(3x-5)-1\right)^(4x+7) $$

Slid vrahuvat, scho so add loneliness is not possible. As if to add loneliness to my embarrassment, then it is necessary and visible, so as not to change the meaning of the whole viraz. For the continuation of the solution, it’s safe that

$$ \frac(3x+1)(3x-5)-1 =\frac(3x+1)(3x-5)-\frac(3x-5)(3x-5) =\frac(3x+1- 3x+5)(3x-5) = frac(6)(3x-5). $$

$\frac(3x+1)(3x-5)-1=\frac(6)(3x-5)$, then:

$$ \lim_(x\to\infty)\left(1+ \frac(3x+1)(3x-5)-1\right)^(4x+7) =\lim_(x\to\infty)\ left(1+\frac(6)(3x-5)\right)^(4x+7) $$

Let's keep going "pidganyannya". The virazi $1+\frac(1)(x)$ formula in the number book contains 1, and in our virazi $1+\frac(6)(3x-5)$ the number book contains $6$. In order to take $1$ from the numeral, let’s drop $6$ from the banner for the help of the offensive transformation:

$$ 1+\frac(6)(3x-5) =1+\frac(1)(\frac(3x-5)(6)) $$

in such a manner,

$$ \lim_(x\to\infty)\left(1+\frac(6)(3x-5)\right)^(4x+7) =\lim_(x\to\infty)\left(1+ \frac(1)(\frac(3x-5)(6))\right)^(4x+7) $$

Father, the basis of the step, tobto. $1+\frac(1)(\frac(3x-5)(6))$, designed to look like $1+\frac(1)(x)$, which is necessary for the formula. Now we can pochnemo pratsyuvati іz pokanik step. Respect what the formula has to say, like the showman of the step and the bannerman, however:

Later, the butt-marker of the step and the banner must be brought to the same form. To take into account the step of the viraz $\frac(3x-5)(6)$, simply multiply the indicator of the step by the whole fraction. Naturally, in order to compensate for such a multiplication, it is possible to multiply it by a good amount, tobto. on $frac(6)(3x-5)$. Father, please:

$$ \lim_(x\to\infty)\left(1+\frac(1)(\frac(3x-5)(6))\right)^(4x+7) =\lim_(x\to\ ) infty)\left(1+\frac(1)(\frac(3x-5)(6))\right)^(\frac(3x-5)(6)\cdot\frac(6)(3x- 5 )\cdot(4x+7)) =\lim_(x\to\infty)\left(\left(1+\frac(1)(\frac(3x-5)(6))\right)^( \ frac(3x-5)(6))\right)^(\frac(6\cdot(4x+7))(3x-5)) $$

Let's take a closer look at the fraction $\frac(6\cdot(4x+7))(3x-5)$, rolled in steps:

$$ \lim_(x\to\infty)\frac(6\cdot(4x+7))(3x-5) =\left|\frac(\infty)(\infty)\right| =\lim_(x\to\infty)\frac(6\cdot\left(4+\frac(7)(x)\right))(3-\frac(5)(x)) =6\cdot\ frac(4)(3)=8. $$

Vidpovid: $\lim_(x\to(0))\biggl(\cos(2x)\biggr)^(\frac(1)(\sin^2(3x)))=e^(-\frac(2) (9)) $.

Butt #4

Find between $\lim_(x\to+\infty)x\left(\ln(x+1)-\ln(x)\right)$.

Shards for $x>0$ can be $\ln(x+1)-\ln(x)=\ln\left(\frac(x+1)(x)\right)$, then:

$$ \lim_(x\to+\infty)x\left(\ln(x+1)-\ln(x)\right) =\lim_(x\to+\infty)\left(x\cdot\ln\ left(\frac(x+1)(x)\right)\right) $$

Expanding the fractions $\frac(x+1)(x)$ into the sum of the fractions $\frac(x+1)(x)=1+\frac(1)(x)$ we take:

$$ \lim_(x\to+\infty)\left(x\cdot\ln\left(\frac(x+1)(x)\right)\right) =\lim_(x\to+\infty)\left (x\cdot\ln\left(1+\frac(1)(x)\right)\right) =\lim_(x\to+\infty)\left(\ln\left(\frac(x+1) ) (x)\right)^x\right) =\ln(e) =1. $$

Vidpovid: $\lim_(x\to+\infty)x\left(\ln(x+1)-\ln(x)\right)=1$.

Butt #5

Find between $\lim_(x\to(2))\biggl(3x-5\biggr)^(\frac(2x)(x^2-4))$.

Shards $\lim_(x\to(2))(3x-5)=6-5=1$ i $\lim_(x\to(2))\frac(2x)(x^2-4)= \ infty$, we can rightly see $1^\infty$. Detailed explanations are given in butt No. 2, immediately mixed with short solutions. After replacing $ t = x-2 $, we take:

$$ \lim_(x\to(2))\biggl(3x-5\biggr)^(\frac(2x)(x^2-4)) =\left|\begin(aligned)&t=x-2 ;\;x=t+2\\t(0)\end(aligned)\right| =\lim_(t\to(0))\biggl(1+3t\biggr)^(\frac(2t+4)(t^2+4t))=\\ =\lim_(t\to(0) )\biggl(1+3t\biggr)^(\frac(1)(3t)\cdot 3t\cdot\frac(2t+4)(t^2+4t)) =\lim_(t\to(0) )\left(\biggl(1+3t\biggr)^(\frac(1)(3t))\right)^(\frac(6\cdot(t+2))(t+4)) =e^ 3. $$

You can untie the butt and also, vicarist change: $t=\frac(1)(x-2)$. Understanding, you will be yourself:

$$ \lim_(x\to(2))\biggl(3x-5\biggr)^(\frac(2x)(x^2-4)) =\left|\begin(aligned)&t=\frac( 1) (x-2); x; =\lim_(t\to\infty)\left(1+\frac(3)(t)\right)^(t\cdot\frac(4t+2)(4t+1))=\\ =\lim_ (t\to\infty)\left(1+\frac(1)(\frac(t)(3))\right)^(\frac(t)(3)\cdot\frac(3)(t) \cdot\frac(t\cdot(4t+2))(4t+1)) =\lim_(t\to\infty)\left(\left(1+\frac(1)(\frac(t)( 3))\right)^(\frac(t)(3))\right)^(\frac(6\cdot(2t+1))(4t+1)) =e^3. $$

Vidpovid: $\lim_(x\to(2))\biggl(3x-5\biggr)^(\frac(2x)(x^2-4))=e^3$.

Butt #6

Find between $\lim_(x\to\infty)\left(\frac(2x^2+3)(2x^2-4)\right)^(3x) $.

It's obvious why $\frac(2x^2+3)(2x^2-4)$ wasted $x\to\infty$:

$$ \lim_(x\to\infty)\frac(2x^2+3)(2x^2-4) =\left|\frac(\infty)(\infty)\right| =\lim_(x\to\infty)\frac(2+\frac(3)(x^2))(2-\frac(4)(x^2)) =\frac(2+0)(2 -0) = 1. $$

In this order, at the task of the boundary, it is possible to the right with the non-insignificance of the mind $1^\infty$, as if rozkriєmo for the help of another miraculous boundary:

$$ \lim_(x\to\infty)\left(\frac(2x^2+3)(2x^2-4)\right)^(3x) =|1^\infty| =\lim_(x\to\infty)\left(1+\frac(2x^2+3)(2x^2-4)-1\right)^(3x)=\\ =\lim_(x\to \infty)\left(1+\frac(7)(2x^2-4)\right)^(3x) =\lim_(x\to\infty)\left(1+\frac(1)(\frac (2x^2-4)(7))\right)^(3x)=\\ =\lim_(x\to\infty)\left(1+\frac(1)(\frac(2x^2-4 ) )(7))\right)^(\frac(2x^2-4)(7)\cdot\frac(7)(2x^2-4)\cdot 3x) =\lim_(x\to\infty ) \left(\left(1+\frac(1)(\frac(2x^2-4)(7))\right)^(\frac(2x^2-4)(7))\right)^ ( \frac(21x)(2x^2-4)) =e^0 =1. $$

Vidpovid: $\lim_(x\to\infty)\left(\frac(2x^2+3)(2x^2-4)\right)^(3x)=1$.

What is the boundary? Understanding the boundaries

Everyone without blame here at the depths of the soul will understand what is the boundary, but only they can smell “between functions” or “between succession”, blame for easy destruction.

Don't fight, you don't know anymore! After 3 hours of reading the lower written, you will become literate.

It is important once and for all to understand what to think about, if you are talking about borderline positions, meanings, situations and vzagali, if you go into life to the term boundary.

Older people think intuitively, but we understand on a few butts.

butt first

Let's guess the rows from the songs of the "Chayf" group: "... do not bring to the edge, do not bring to the edge ...".

butt other

Without a doubt, forget the phrase about the borderline stand of the object in space.

You yourself can easily simulate such a situation with handy speeches.

For example, lightly heal plastic bowl and let її. Vaughn turn around on the bottom.

Ale є є such borderline pohili position, beyond the borders of these, it’s just a fall.

Well, I know the border camp in to this particular type- Be more specific. It is important to think.

It is possible to suggest a rich application of the term boundary: between human capabilities, between material worth, then.

Well, with the lawless, we are so excited today)))

But at the same time, we are to be called between sequences and between functions in mathematics.

Between numerical sequences in mathematics

Boundary (numerical sequence) - one of the main things to understand mathematical analysis. Hundreds and hundreds of theorems are based on the understanding of the boundary transition, which signify modern science.

See a specific example for accuracy.

It is permissible to have an inexhaustible sequence of numbers, the skin of which is half the size of the front, starting from one: 1, ½, ¼, ...

So the axis between the numerical sequence (as it is not true) is, however, more specific.

In the course of the process, the skin approached the significance of the sequence, it was not close to the song number.

It doesn't matter if you guess that it will be zero.

Important!

If we are talking about the basis of the boundary (boundary value), it does not mean that any member of the sequence is more valuable to the boundary value. Vіn can only pragnet to the new.

From our butt, I realized more. Skіlki brazіv mi did not divide one by two, mi never take zero. If the number of two is less than in front, but not zero!

Interfacial functions in mathematics

In mathematical analysis, insanely, more importantly - ce understanding between functions.

Without going deep into the theory, let's say this: the boundary value of a function, which can always lie within the scope of the value of the function itself.

When the argument is changed, the function will accept any value, or it may not be accepted at all.

For example, hyperbole 1/x do not have a value of zero at the same point, but it is not set to zero at the right point x to inexcusability.

Calculator between

Our method cannot give you some theoretical knowledge, for which there are impersonal sensible books.

Ale mi proponuєmo you speed up online calculator between, for the help of someone you can equalize your decision with the correct opinion.

All in all, the calculator looks like a partial solution between, zastosovuyuchi most often Lopital's rule with different differentiation of the numerator and the banner without interruption in the point or on the other side of the function.

The first miraculous boundary is called such equivalence:

\begin(equation)\lim_(\alpha\to(0))\frac(\sin\alpha)(\alpha)=1 \end(equation)

So if $ \ alpha \ to (0) $ can be $ \ sin \ alpha \ to (0) $, then it seems that the first miracle of the boundary between the curves is indistinct in the form of $ \ frac (0) (0) $. Seemingly, in the formula (1) the replacement of the changing $ \ alpha $ under the sign of the sine і in the banner can be ruffled, be it an expression, - two minds were abyssed:

1. Vyslovlyuvannya under the sign of the sine and at the sign of the standard one hour to jump zero, tobto. є insignificance of the form $\frac(0)(0)$.
2. Virazi under the sign of the sine and the standard sign run.

Often there are also traces from the first miraculous boundary:

\begin(equation) \lim_(\alpha\to(0))\frac(\tg\alpha)(\alpha)=1 \end(equation) \begin(equation) \lim_(\alpha\to(0) )\frac(\arcsin\alpha)(\alpha)=1 \end(equation) \begin(equation) \lim_(\alpha\to(0))\frac(\arctg\alpha)(\alpha)=1 \end(equation)

Eleven butts are written on the third side. Butt No. 1 assignments to the proof of formulas (2) - (4). Apply No. 2, No. 3, No. 4 and No. 5 to answer the decision with the report comments. Apply No. 6-10 to the decision practically without comments, but the report of the explanation was given at the front butts. When victorious, there are some trigonometric formulas, which you can know.

I respect that the presence of trigonometric functions together with the non-insignificance of $\frac(0)(0)$ does not yet mean the obsession of the first miraculous boundary. Sometimes you can complete simple trigonometric transformations, for example, divas.

Butt #1

Bring what $\lim_(\alpha\to(0))\frac(\tg\alpha)(\alpha)=1$, $\lim_(\alpha\to(0))\frac(\arcsin\alpha )(\alpha)=1$, $\lim_(\alpha\to(0))\frac(\arctg\alpha)(\alpha)=1$.

a) Since $\tg\alpha = \frac (\sin\alpha)(\cos\alpha)$, then:

$$ \lim_(\alpha\to(0))\frac(\tg(\alpha))(\alpha)=\left|\frac(0)(0)\right| =\lim_(\alpha\to(0))\frac(\sin(\alpha))(\alpha\cos(\alpha)) $$

Skіlki $\lim_(\alpha\to(0))\cos(0)=1$ і $\lim_(\alpha\to(0))\frac(\sin\alpha)(\alpha)=1$ , then:

$$ \lim_(\alpha\to(0))\frac(\sin(\alpha))(\alpha\cos(\alpha)) =\frac(\displaystyle\lim_(\alpha\to(0)) \frac(\sin(\alpha))(\alpha))(\displaystyle\lim_(\alpha\to(0))\cos(\alpha)) =\frac(1)(1) =1. $$

b) I will replace $ \ alpha = \ sin (y) $. If $\sin(0)=0$, then think $\alpha\to(0)$ maybe $y\to(0)$. In addition, іsnuє around zero, in the same $\arcsin\alpha=\arcsin(\sin(y))=y$, to that:

$$ \lim_(\alpha\to(0))\frac(\arcsin\alpha)(\alpha)=\left|\frac(0)(0)\right| =\lim_(y\to(0))\frac(y)(\sin(y)) =\lim_(y\to(0))\frac(1)(\frac(\sin(y))( y)) =\frac(1)(\displaystyle\lim_(y\to(0))\frac(\sin(y))(y)) =\frac(1)(1) =1. $$

$\lim_(\alpha\to(0))\frac(\arcsin\alpha)(\alpha)=1$ completed.

c) Let me change $ alpha = tg (y) $. If $\tg(0)=0$, then think $\alpha\to(0)$ and $y\to(0)$ are equivalent. In addition, if you base around zero, in such a way $\arctg\alpha=\arctg\tg(y))=y$, then, relying on the results of point a), we can:

$$ \lim_(\alpha\to(0))\frac(\arctg\alpha)(\alpha)=\left|\frac(0)(0)\right| =\lim_(y\to(0))\frac(y)(\tg(y)) =\lim_(y\to(0))\frac(1)(\frac(\tg(y))( y)) =\frac(1)(\displaystyle\lim_(y\to(0))\frac(\tg(y))(y)) =\frac(1)(1) =1. $$

$\lim_(\alpha\to(0))\frac(\arctg\alpha)(\alpha)=1$ completed.

Rivnosti a), b), c) are often victorious in order from the first miraculous boundary.

Butt #2

Calculate between $\lim_(x\to(2))\frac(\sin\left(\frac(x^2-4)(x+7)\right))(\frac(x^2-4)( x+7))$.

Scales $\lim_(x\to(2))\frac(x^2-4)(x+7)=\frac(2^2-4)(2+7)=0$ i $\lim_( x \to(2))\sin\left(\frac(x^2-4)(x+7)\right)=\sin(0)=0$, so. If the numerator and the banner of the fraction immediately go to zero, then it can be right with the non-significance of the form $\frac(0)(0)$, then. viconano. In addition, it is clear that virazi under the sign of the sine i in the banner are running (tobto vikonana i):

Otzhe, offended mind, transferred to the cob of the side, vikonan. On tsimu vyplivaє, scho zastosovna formula, tobto. $\lim_(x\to(2)) \frac(\sin\left(\frac(x^2-4)(x+7)\right))(\frac(x^2-4)(x+ 7 )) = $1.

Vidpovid: $\lim_(x\to(2))\frac(\sin\left(\frac(x^2-4)(x+7)\right))(\frac(x^2-4)(x +7)) = $1.

Butt #3

Know $\lim_(x\to(0))\frac(\sin(9x))(x)$.

Scales $\lim_(x\to(0))\sin(9x)=0$ і $\lim_(x\to(0))x=0$, but we can use $\frac(0 ) ( 0) $, then. viconano. Prote virazi under the sign of sine and standard do not escape. Here it is necessary to give a viraz to the bannerman in the necessary form. It is necessary for us, if the flagman has a roztashuvavsya $9x$ - then you will become true. As a matter of fact, we don't get a multiplier of $9$ from the bannerman, which is not so easy to introduce - just multiply the viraz from the bannerman by $9$. Naturally, to compensate for the multiplication of $9$, you need to multiply by $9$ and divide:

$$ \lim_(x\to(0))\frac(\sin(9x))(x)=\left|\frac(0)(0)\right| =\lim_(x\to(0))\frac(\sin(9x))(9x\cdot\frac(1)(9)) =9\lim_(x\to(0))\frac(\sin (9x))(9x) $$

Now, at the banner, that under the sign of the sine, they scurried. Wash your mind for the inter $\lim_(x\to(0))\frac(\sin(9x))(9x)$ vikonanі. Also, $\lim_(x\to(0))\frac(\sin(9x))(9x)=1$. And tse means that:

$$ 9\lim_(x\to(0))\frac(\sin(9x))(9x)=9cdot(1)=9. $$

Vidpovid: $\lim_(x\to(0))\frac(\sin(9x))(x)=9$.

Butt #4

Know $\lim_(x\to(0))\frac(\sin(5x))(\tg(8x))$.

$\lim_(x\to(0))\sin(5x)=0$ і $\lim_(x\to(0))\tg(8x)=0$, then we can rightly see that $ \frac(0)(0)$. However, the form of the first miraculous boundary is broken. The chiselnik, which avenges $\sin(5x)$, means the presence of the banner $5x$. In this situation, it is easiest to divide the number by $5x$, - and multiply by $5x$. In addition, it is probable that the operation is similar to the one with the standard, multiplying it and dividing $\tg(8x)$ by $8x$:

$$\lim_(x\to(0))\frac(\sin(5x))(\tg(8x))=\left|\frac(0)(0)\right| =\lim_(x\to(0))\frac(\frac(\sin(5x))(5x)\cdot(5x))(\frac(\tg(8x))(8x)\cdot(8x) )$$

If the constant $\frac(5)(8)$ is fast on $x$ i, then we take:

$$ \lim_(x\to(0))\frac(\frac(\sin(5x))(5x)\cdot(5x))(\frac(\tg(8x))(8x)\cdot(8x )) =\frac(5)(8)\cdot\lim_(x\to(0))\frac(\frac(\sin(5x))(5x))(\frac(\tg(8x))( 8x)) $$

Respect that $\lim_(x\to(0))\frac(\sin(5x))(5x)$ is more than happy for the first wonderland. For $\lim_(x\to(0))\frac(\tg(8x))(8x)$, the stagnant formula is:

$$ \frac(5)(8)\cdot\lim_(x\to(0))\frac(\frac(\sin(5x))(5x))(\frac(\tg(8x))(8x )) =\frac(5)(8)\cdot\frac(\displaystyle\lim_(x\to(0))\frac(\sin(5x))(5x))(\displaystyle\lim_(x\to (0))\frac(\tg(8x))(8x)) =\frac(5)(8)\cdot\frac(1)(1) =\frac(5)(8). $$

Vidpovid: $\lim_(x\to(0))\frac(\sin(5x))(\tg(8x))=\frac(5)(8)$.

Butt #5

Know $\lim_(x\to(0))\frac(\cos(5x)-\cos^3(5x))(x^2)$.

Skіlki $\lim_(x\to(0))(\cos(5x)-\cos^3(5x))=1-1=0$ (guess that $\cos(0)=1$) and $ \lim_(x\to(0))x^2=0$; However, in order to zastosuvat the first miraculous boundary, slid the cosine into the number book, moving to sines (so that we will zastosuvat the formula) or tangents (so we will zastosuvat the formula). Zrobiti tse can be such transformations:

$$\cos(5x)-\cos^3(5x)=\cos(5x)\cdot\left(1-\cos^2(5x)\right)$$ $$\cos(5x)-\cos ^3(5x)=\cos(5x)\cdot\left(1-\cos^2(5x)\right)=\cos(5x)\cdot\sin^2(5x).$$

Let's turn to the boundary:

$$ \lim_(x\to(0))\frac(\cos(5x)-\cos^3(5x))(x^2)=\left|\frac(0)(0)\right| =\lim_(x\to(0))\frac(\cos(5x)\cdot\sin^2(5x))(x^2) =\lim_(x\to(0))\left(\cos (5x)\cdot\frac(\sin^2(5x))(x^2)\right) $$

The fraction $\frac(\sin^2(5x))(x^2)$ is already close to that form, which is necessary for the first miraculous boundary. Trochs are corrected with the fraction $\frac(\sin^2(5x))(x^2)$, pіdganyayuchi її pіd pershu miraculous boundary (fuck, scho vrazi in the number book and pіd sine due to zbіgtisya):

$$\frac(\sin^2(5x))(x^2)=\frac(\sin^2(5x))(25x^2\cdot\frac(1)(25))=25\cdot\ frac(\sin^2(5x))(25x^2)=25\cdot\left(\frac(\sin(5x))(5x)\right)^2$$

Let's turn to the boundary:

$$ \lim_(x\to(0))\left(\cos(5x)\cdot\frac(\sin^2(5x))(x^2)\right) =\lim_(x\to(0 ) ))\left(25\cos(5x)\cdot\left(\frac(\sin(5x))(5x)\right)^2\right)=\=25\cdot\lim_(x\to( 0))\cos(5x)\cdot\lim_(x\to(0))\left(\frac(\sin(5x))(5x)\right)^2 =25\cdot(1)\cdot( 1^2) = 25.$$

Vidpovid: $\lim_(x\to(0))\frac(\cos(5x)-\cos^3(5x))(x^2)=25$.

Butt #6

Find between $\lim_(x\to(0))\frac(1-\cos(6x))(1-\cos(2x))$.

Scales $\lim_(x\to(0))(1-\cos(6x))=0$ і $\lim_(x\to(0))(1-\cos(2x))=0$, mi maybe right from the non-insignificance of $\frac(0)(0)$. Rozkriёmo її for the help of the first miraculous boundary. For which we pass from cosines to sines. $1-\cos(2\alpha)=2\sin^2(\alpha)$, then:

$$1-\cos(6x)=2\sin^2(3x);\;1-\cos(2x)=2\sin^2(x).$$

Passing in the task between to sinuses, matimemo:

$$ \lim_(x\to(0))\frac(1-\cos(6x))(1-\cos(2x))=\left|\frac(0)(0)\right| =\lim_(x\to(0))\frac(2\sin^2(3x))(2\sin^2(x)) =\lim_(x\to(0))\frac(\sin^ 2(3x))(\sin^2(x))=\\ =\lim_(x\to(0))\frac(\frac(\sin^2(3x))((3x)^2)\ cdot(3x)^2)(\frac(\sin^2(x))(x^2)\cdot(x^2)) =\lim_(x\to(0))\frac(\left(\ ) frac(\sin(3x))(3x)\right)^2\cdot(9x^2))(\left(\frac(\sin(x))(x)\right)^2\cdot(x ^ 2)) =9\cdot\frac(\displaystyle\lim_(x\to(0))\left(\frac(\sin(3x))(3x)\right)^2)(\displaystyle\lim_( x \to(0))\left(\frac(\sin(x))(x)\right)^2) =9cdot\frac(1^2)(1^2) =9. $$

Vidpovid: $\lim_(x\to(0))\frac(1-\cos(6x))(1-\cos(2x))=9$.

Butt #7

Calculate between $\lim_(x\to(0))\frac(\cos(\alpha(x))-\cos(\beta(x)))(x^2)$ for $\alpha\neq\ beta$.

Detailed explanations were given earlier, here it is simply significant that $\frac(0)(0)$ is again insignificant. Let's move on from cosines to sines, victorious formula

$$\cos\alpha-\cos\beta=-2\sin\frac(\alpha+\beta)(2)\cdot\sin\frac(\alpha-\beta)(2).$$

The vicorist formula is shown, it is necessary:

$$ \lim_(x\to(0))\frac(\cos(\alpha(x))-\cos(\beta(x)))(x^2)=\left|\frac(0)( 0) \right| =\lim_(x\to(0))\frac(-2\sin\frac(\alpha(x)+\beta(x))(2)\cdot\sin\frac(\alpha(x)-\ beta(x))(2))(x^2)=\\=-2\cdot\lim_(x\to(0))\frac(\sin\left(x\cdot\frac(\alpha+\beta ) )(2)\right)\cdot\sin\left(x\cdot\frac(\alpha-beta)(2)\right))(x^2) =-2\cdot\lim_(x\to( 0))\left(\frac(\sin\left(x\cdot\frac(\alpha+\beta)(2)\right))(x)\cdot\frac(\sin\left(x\cdot\frac ) (\alpha-\beta)(2)\right))(x)\right)=\\ =-2\cdot\lim_(x\to(0))\left(\frac(\sin\left( x) \cdot\frac(\alpha+\beta)(2)\right))(x\cdot\frac(\alpha+\beta)(2))\cdot\frac(\alpha+\beta)(2)\cdot \frac (\sin\left(x\cdot\frac(\alpha-\beta)(2)\right))(x\cdot\frac(\alpha-\beta)(2))\cdot\frac(\ alpha- \beta)(2)\right)=\\=-\frac((\alpha+\beta)\cdot(\alpha-\beta))(2)\lim_(x\to(0))\frac (\ sin\left(x\cdot\frac(\alpha+\beta)(2)\right))(x\cdot\frac(\alpha+\beta)(2))\cdot\lim_(x\to(0 )) \frac(\sin\left(x\cdot\frac(\alpha-\beta)(2)\right))(x\cdot\frac(\alpha-\beta)(2)) =-\frac (\alpha^2-\beta^2)(2)\cdot(1)\cdot(1) =\frac(\beta^2-\alpha^2)(2). $$

Vidpovid: $\lim_(x\to(0))\frac(\cos(\alpha(x))-\cos(\beta(x)))(x^2)=\frac(\beta^2-\ alpha^2) (2) $.

Butt #8

Find between $\lim_(x\to(0))\frac(\tg(x)-\sin(x))(x^3)$.

$\lim_(x\to(0))(\tg(x)-\sin(x))=0$ (guess $\sin(0)=\tg(0)=0$) and $ \lim_(x\to(0))x^3=0$, then we can right-hand with the non-insignificance of the form $\frac(0)(0)$. Rozkriemo її like this:

$$ \lim_(x\to(0))\frac(\tg(x)-\sin(x))(x^3)=\left|\frac(0)(0)\right| =\lim_(x\to(0))\frac(\frac(\sin(x))(\cos(x))-\sin(x))(x^3) =\lim_(x\to( 0))\frac(\sin(x)\cdot\left(\frac(1)(\cos(x))-1\right))(x^3) =\lim_(x\to(0)) \frac(\sin(x)\cdot\left(1-\cos(x)\right))(x^3\cdot\cos(x))=\\ =\lim_(x\to(0)) \frac(\sin(x)\cdot(2)\sin^2\frac(x)(2))(x^3\cdot\cos(x)) =\frac(1)(2)\cdot\ lim_(x\to(0))\left(\frac(\sin(x))(x)\cdot\left(\frac(\sin\frac(x)(2))(\frac(x)( 2))\right)^2\cdot\frac(1)(\cos(x))\right) =\frac(1)(2)\cdot(1)\cdot(1^2)\cdot(1 ) = frac(1)(2). $$

Vidpovid: $\lim_(x\to(0))\frac(\tg(x)-\sin(x))(x^3)=\frac(1)(2)$.

Butt #9

Find between $\lim_(x\to(3))\frac(1-\cos(x-3))((x-3)\tg\frac(x-3)(2))$.

Scales $\lim_(x\to(3))(1-\cos(x-3))=0$ і $\lim_(x\to(3))(x-3)\tg\frac(x - 3) (2) = 0 $, then $ \ frac (0) (0) $ does not exist. Before that, as you go to the її opening, manually change the change in such a rank, so that the new change is straightened to zero (reveal that the formulas change $\alpha\to 0$). It's easier to enter the change $t=x-3$. However, for the convenience of the distant transformations (you can always remember the hour of the decision below), you can change this change: $t=\frac(x-3)(2)$. I’ll appoint that I offended you by replacing zastosovnі in this particular situation, just to allow a friend to replace him less with fractions. $x\to(3)$, then $t\to(0)$.

$$ \lim_(x\to(3))\frac(1-\cos(x-3))((x-3)\tg\frac(x-3)(2))=\left|\frac (0)(0)\right| =\left|\begin(aligned)&t=\frac(x-3)(2);\&t\to(0)\end(aligned)\right| =\lim_(t\to(0))\frac(1-\cos(2t))(2t\cdot\tg(t)) =\lim_(t\to(0))\frac(2\sin^ 2t)(2t\cdot\tg(t)) =\lim_(t\to(0))\frac(\sin^2t)(t\cdot\tg(t))=\\ =\lim_(t\ to(0))\frac(\sin^2t)(t\cdot\frac(\sin(t))(\cos(t))) =\lim_(t\to(0))\frac(\sin (t)\cos(t))(t) =\lim_(t\to(0))\left(\frac(\sin(t))(t)\cdot\cos(t)\right) =\ lim_(t\to(0))\frac(\sin(t))(t)\cdot\lim_(t\to(0))\cos(t) =1\cdot(1) =1. $$

Vidpovid: $\lim_(x\to(3))\frac(1-\cos(x-3))((x-3)\tg\frac(x-3)(2))=1$.

Butt #10

Find between $\lim_(x\to\frac(\pi)(2))\frac(1-\sin(x))(\left(\frac(\pi)(2)-x\right)^2 ) $.

May be I can renew from the right $\frac(0)(0)$. Before that, as you go to the її opening, manually change the change in such a rank, so that the new change is straightened to zero (respect that the formulas change $\alpha\to(0)$). The easiest way is to enter change $t=\frac(\pi)(2)-x$. $x\to\frac(\pi)(2)$, then $t\to(0)$:

$$ \lim_(x\to\frac(\pi)(2))\frac(1-\sin(x))(\left(\frac(\pi)(2)-x\right)^2) =\left|\frac(0)(0)\right| =\left|\begin(aligned)&t=\frac(\pi)(2)-x;\&t\to(0)\end(aligned)\right| =\lim_(t\to(0))\frac(1-\sin\left(\frac(\pi)(2)-t\right))(t^2) =\lim_(t\to(0 ) ))\frac(1-\cos(t))(t^2)=\\ =\lim_(t\to(0))\frac(2\sin^2\frac(t)(2)) ( t^2) =2\lim_(t\to(0))\frac(\sin^2\frac(t)(2))(t^2) =2\lim_(t\to(0)) \frac(\sin^2\frac(t)(2))(\frac(t^2)(4)\cdot(4)) =\frac(1)(2)\cdot\lim_(t\to ( 0))\left(\frac(\sin\frac(t)(2))(\frac(t)(2))\right)^2 =\frac(1)(2)\cdot(1^ 2 ) = frac(1)(2). $$

Vidpovid: $\lim_(x\to\frac(\pi)(2))\frac(1-\sin(x))(\left(\frac(\pi)(2)-x\right)^2) = frac(1)(2)$.

Stock #11

Find between $\lim_(x\to\frac(\pi)(2))\frac(1-\sin(x))(\cos^2x)$, $\lim_(x\to\frac(2\ )pi)(3))\frac(\tg(x)+\sqrt(3))(2\cos(x)+1)$.

We won’t be able to vicorate the first miraculous boundary at this vipadk. To give respect: as in the first, so in the other boundary there are only trigonometric functions of that number. Most often in such butts one can say to prostit viraz, roztashovane under the sign of the boundary. With the help of a guessed forgiveness, that quickness of the deaky spіvmulnіnіnіnіnіnіnіnіnіє znikає. I’ve created this butt only with one method: show that the presence of trigonometric functions under the sign of the boundary does not necessarily mean that the first miraculous boundary is stuck.

Skіlki $\lim_(x\to\frac(\pi)(2))(1-\sin(x))=0$ (guess $\sin\frac(\pi)(2)=1$ ) i $\lim_(x\to\frac(\pi)(2))\cos^2x=0$ (guess that $\cos\frac(\pi)(2)=0$), then we can out of insignificance $ frac (0) (0) $. However, tse zovsіm does not mean that it is necessary for us to conquer the first miraculous boundary. To reveal the insignificance, tell the truth about $\cos^2x=1-\sin^2x$:

$$ \lim_(x\to\frac(\pi)(2))\frac(1-\sin(x))(\cos^2x) =\left|\frac(0)(0)\right| =\lim_(x\to\frac(\pi)(2))\frac(1-\sin(x))(1-\sin^2x) =\lim_(x\to\frac(\pi)( 2))\frac(1-\sin(x))((1-\sin(x))(1+\sin(x))) =\lim_(x\to\frac(\pi)(2) )\frac(1)(1+\sin(x)) = frac(1)(1+1) = frac(1)(2). $$

Analogous method of solution Reshnik Demidovich (No. 475). As far as the other boundary, those as in the front butts of which I divided, we may not see $\frac(0)(0)$. Why is she blaming? It's because $ \ tg \ frac (2 \ pi) (3) = - \ sqrt (3) $ i $ 2 \ cos \ frac (2 \ pi) (3) = -1 $. Vikoristovuєmo tsі meaning with the method of transformation of virazіv at the numeral and at the bannerman. The meta of our actions: write down the sum in the number book and the banner in front of the creation. Before the speech, often in the boundaries of a similar kind, the replacement of the change is chuckled, with such a rose, so that the new change is straightened to zero (div., for example, butt No. 9 or No. 10 on the other side). However, this butt has no sense in replacing the sensor, wanting to change the $t=x-\frac(2\pi)(3)$ for the bazhanya is clumsy.

$$ \lim_(x\to\frac(2\pi)(3))\frac(\tg(x)+\sqrt(3))(2\cos(x)+1) =\lim_(x\ to\frac(2\pi)(3))\frac(\tg(x)+\sqrt(3))(2\cdot\left(\cos(x)+\frac(1)(2)\right )) =\lim_(x\to\frac(2\pi)(3))\frac(\tg(x)-\tg\frac(2\pi)(3))(2\cdot\left(\ ) cos(x)-\cos\frac(2\pi)(3)\right))=\\ =\lim_(x\to\frac(2\pi)(3))\frac(\frac(\ sin) \left(x-\frac(2\pi)(3)\right))(\cos(x)\cos\frac(2\pi)(3)))(-4\sin\frac(x+ \frac) (2\pi)(3))(2)\sin\frac(x-\frac(2\pi)(3))(2)) =\lim_(x\to\frac(2\pi )(3 ))\frac(\sin\left(x-\frac(2\pi)(3)\right))(-4\sin\frac(x+\frac(2\pi)(3))( 2)\ sin\frac(x-\frac(2\pi)(3))(2)\cos(x)\cos\frac(2\pi)(3)) =\\ =\lim_(x\ to\frac (2\pi)(3))\frac(2\sin\frac(x-\frac(2\pi)(3))(2)\cos\frac(x-frac(2\pi) (3) ))(2))(-4\sin\frac(x+\frac(2\pi)(3))(2)\sin\frac(x-\frac(2\pi)(3)) (2) \cos(x)\cos\frac(2\pi)(3)) =\lim_(x\to\frac(2\pi)(3))\frac(\cos\frac(x-\ frac(2) \pi)(3))(2))(-2\sin\frac(x+\frac(2\pi)(3))(2)\cos(x)\cos\frac(2\ pi)(3 ))=\\ =\frac(1)(-2\cdot\frac(\sqrt(3))(2)\cdot\left(-\frac(1)(2)\right)\ cdot\left( -\frac(1)(2)\right)) =-\frac(4 )(\sqrt(3)). $$

Yak bachite, we didn’t have a chance to zastosovuvat Persh miraculous boundary. Zvichayno, for bazhannya tse you can rob (div. note below), but you can’t consume it.

What will be the solution to the first miracle of the miraculous boundary? show/hide

With the victory of the first miraculous boundary, it is necessary:

$$ \lim_(x\to\frac(2\pi)(3))\frac(\sin\left(x-\frac(2\pi)(3)\right))(-4\sin\frac (x+\frac(2\pi)(3))(2)\sin\frac(x-frac(2\pi)(3))(2)\cos(x)\cos\frac(2\pi ) (3))=\\ =\lim_(x\to\frac(2\pi)(3))\left(\frac(\sin\left(x-frac(2\pi)(3)\) right ))(x-\frac(2\pi)(3))\cdot\frac(1)(\frac(\sin\frac(x-frac(2\pi)(3))(2)) (\ frac(x-\frac(2\pi)(3))(2)))\cdot\frac(1)(-2\sin\frac(x+\frac(2\pi)(3))( 2) \cos(x)\cos\frac(2\pi)(3))\right) =1cdot(1)cdotfrac(1)(-2cdotfrac(sqrt(3)) )(2)\cdot\left(-\ frac(1)(2)\right)\cdot\left(-\frac(1)(2)\right)) =-\frac(4)(\sqrt( 3)). $$

Vidpovid: $\lim_(x\to\frac(\pi)(2))\frac(1-\sin(x))(\cos^2x)=\frac(1)(2)$, $\lim_( x\to\frac(2\pi)(3))\frac(\tg(x)+\sqrt(3))(2\cos(x)+1)=-\frac(4)(\sqrt( 3)) $.

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