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How do I ind the limit of

$$\lim_{x \rightarrow 0}\frac{\tan x- x-\frac{x^3}{3}}{\sin^5x}$$

by L'Hopital's Rule?

Using Desmos, I get the answer that this limit evaluates to $\frac{3}{15}$, but I can't get that answer.

This is what I've done so far:

\begin{align} &\lim_{x \rightarrow 0}\frac{\tan x -x}{\sin^5x}-\lim_{x \rightarrow 0}\frac{x^3/3}{\sin^5x}\\ &=\lim_{x \rightarrow 0}\frac{\sec^2x-1}{5\sin^4x\cos x}-\lim_{x \rightarrow 0}\frac{x^2}{5\sin^4x\cos x}\\ &=\lim_{x \rightarrow 0}\frac{\tan^2x}{5\sin^4x\cos x}-\lim_{x \rightarrow 0}\frac{x^2}{\sin^2x}\lim_{x \rightarrow 0}\frac{1}{5\sin^2x\cos x}\\ &=\lim_{x \rightarrow 0}\frac{1}{5\sin^2x\cos^3x}-\lim_{x \rightarrow 0}\frac{\cos^2x}{5\sin^2x\cos^3 x}\\ &=\frac{1}{5} \end{align}

I checked by answer with Desmos, and the 4th line is where the difference occurs. In the 4th line, I did this.

$$ \lim_{x \rightarrow 0}\frac{x^2}{\sin^2x}=\lim_{x \rightarrow 0}\frac{1}{(\sin x/x)^2}=1 $$

Shouldn't this be a valid operation? I'm just using the product law on limits. If anyone could tell me where I've gone wrong, I will be grateful.

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  • $\begingroup$ Multiply top and bottom by x^5 and split into a product. You get one limit for free $\endgroup$ – imranfat Oct 6 at 14:57
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To begin with, you split the limit into two ones that are $+\infty$, so the very first line is $\infty-\infty$. Even though those two infinities are comparable at the beginning (the limit exists) you cannot handle them separately as numbers and replace $\infty$ with $1\cdot\infty$. Note that all the operations with limits (sum, product, fraction) are justified only when all the limits exists and are finite. Take a simple example $$ \lim_{x\to 0}\frac{\sin x-x}{x^3}=-\frac{1}{6}\quad (\text{L'Hospital two times}), $$ but if you do what you did $$ \lim_{x\to 0}\frac{\sin x-x}{x^3}=\lim_{x\to 0}\frac{\sin x}{x^3}-\lim_{x\to 0}\frac{x}{x^3}=\underbrace{\lim_{x\to 0}\frac{\sin x}{x}}_{=1}\lim_{x\to 0}\frac{1}{x^2}-\lim_{x\to 0}\frac{1}{x^2} $$ you will get nonsense.

In your example, you should keep both fractions under one limit and split them only when you are sure that the limits are finite. You may continue as $$ \lim\frac{\tan^2x-x^2}{5\sin^4x\cos x}=\lim\frac{\tan x-x}{\sin^3 x}\cdot\lim\frac{\tan x+x}{5\sin x\cos x} $$ or rewrite at the beginning as $$ \lim_{x \rightarrow 0}\frac{\tan x- x-\frac{x^3}{3}}{\sin^5x}= \lim_{x \rightarrow 0}\frac{\sin x- x\cos x-\frac{x^3}{3}\cos x}{x^5}\cdot \lim_{x\to 0}\frac{x^5}{\sin^5x\cos x}. $$ The last limit in the product in both cases is easily calculated.

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  • $\begingroup$ For the first case, how can I be sure that both limits: $\lim (tanx-x)/(sin^3x)$ and $\lim (tanx+x)/5sinxcosx$ are finite? $\endgroup$ – Yip Jung Hon Oct 6 at 15:25
  • $\begingroup$ After all, aren't they still indeterminate forms? $\endgroup$ – Yip Jung Hon Oct 6 at 15:31
  • $\begingroup$ @YipJungHon Yes, they are indeterminate, but the second one (nonzero) is relatively easy as both $\frac{\tan x}{\sin x}$ and $\frac{x}{\sin x}$ go to one. So the original limit exists if and only if there exists the first limit $\frac{\tan x-x}{\sin^3 x}$ that one can calculate by L'Hospital rule again. $\endgroup$ – A.Γ. Oct 6 at 16:03
  • $\begingroup$ Thanks for your help. Can I ask: is there a special number to what you've mentioned? That the original limit exists iff there exists the first limit that one can calculated by L'Hospital's Rule? $\endgroup$ – Yip Jung Hon Oct 6 at 23:52
  • $\begingroup$ @YipJungHon What I meant was the following: assume you have a limit $\lim f\cdot g$ that you are not sure to exist, but you know that $\lim g$ exists and is nonzero. Then $\lim f\cdot g$ exists if and only if $\lim f$ exists. Moreover, you can split it as $\lim f\cdot g=\lim f\cdot\lim g$. $\endgroup$ – A.Γ. Oct 7 at 16:32
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As noted by А.Г., you can not split the limit unless both exist.

You can use L'Hospital right away and see how splittings are done: $$\lim_{x \rightarrow 0}\frac{\tan x- x-\frac{x^3}{3}}{\sin^5x}\stackrel{LR}= \lim_{x \rightarrow 0}\frac{\sec ^2x- 1-x^2}{5\sin^4x\cdot \cos x}=\\ \lim_{x \rightarrow 0}\frac{1-\cos^2x-x^2\cos^2x}{5\sin^4x}\cdot \underbrace{\lim_{x\to 0}\frac1{\cos^3x}}_{=1}\stackrel{LR}=\\ \lim_{x \rightarrow 0}\frac{\sin 2x-2x\cos^2x+x^2\sin 2x}{20\sin^3x}\cdot \underbrace{\lim_{x\to 0}\frac{1}{\cos x}}_{=1}\stackrel{LR}=\\ \lim_{x \rightarrow 0}\frac{2\cos 2x-2\cos^2x+4x\sin 2x+2x^2\cos 2x}{60\sin^2x}\cdot \underbrace{\lim_{x\to 0}\frac{1}{\cos x}}_{=1}=\\ \lim_{x \rightarrow 0}\frac{-2\sin^2x}{60\sin^2x}+\lim_{x \rightarrow 0}\frac{4x\sin 2x}{60\sin^2x}+\lim_{x \rightarrow 0}\frac{2x^2\cos 2x}{60\sin^2x}=\\ -\frac1{30}+\frac2{15}+\frac1{30}=\frac2{15}.$$

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  • $\begingroup$ But the answer is supposed to be 3/20 $\endgroup$ – imranfat Oct 6 at 15:51
  • $\begingroup$ Hi sorry, I made a typo, 2/15 is supposed to be the correct answer $\endgroup$ – Yip Jung Hon Oct 6 at 16:07
  • $\begingroup$ You can upvote if this answer was helpful. $\endgroup$ – farruhota Oct 6 at 17:31
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Your approach has a common mistake. The split of limits can't be done unless one of the parts has a finite limit. See more details in this answer.

The limit can be most easily evaluated using Taylor series for $\tan x$. An alternative approach is to use L'Hospital's Rule. First we can replace the denominator by $x^5$ via the use of standard limit $\lim\limits _{x\to 0}\dfrac {\sin x} {x} =1$. And then applying L'Hospital's Rule we get the expression $$\frac{\sec^2x-1-x^2}{5x^4}$$ which can be written as $$\frac{1}{5}\cdot\frac{\tan x - x} {x^3}\cdot\left(1+\frac{\tan x} {x} \right) $$ and therefore the desired limit is equal to $$\frac{2}{5}\lim_{x \to 0}\frac{\tan x - x} {x^3}$$ One can apply L'Hospital's Rule once again and see that the desired limit is $(2/5)(1/3)=2/15$.

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