Unable to evaluate limit correctly

Question

I want to find the limit of $$\frac{e^x + \frac1{e^x} - 2\cos x}{x\tan x}$$ as $x$ tends to $0$.

My attempt: The limit of $\frac{e^x + 1/e^x - 2cosx}{xtanx}$ should be the same as the limit of $\frac{2 - 2\cos x}{x\tan x}$, which can evaluated using the standard limits of $x/\sin x$ and $((1-\cos x)/x^2)$. But this gives me the answer as $1$, while the correct answer is $2$. I suspect the error is in writing it as $\frac{2 - 2cosx}{xtanx}$, but am unable to see why. Please help.

To give context - I can use only the basic limit laws for algebraic combinations of limits and some standard limits.

Basic limit laws for adding, subtracting, multiplying, dividing, etc. — Soumil Aggarwal, Feb 14 '18 at 12:09
No, I haven't learned that yet. It comes later in the textbook. — Soumil Aggarwal, Feb 14 '18 at 12:11
Consider the limit of $$\frac{e^x-e^{-x}+2 \sin x}{\tan x+x \sec^2 x}$$. How did I get here, what could be the next step? — , Feb 14 '18 at 12:11
The textbook does mention using the expansions of $e^x$, $cosx$, $tanx$, but I was wondering if it can be done without them, as no proofs for the former have been given. Can you point out where my method goes wrong and why? — Soumil Aggarwal, Feb 14 '18 at 12:15
@SoumilAggarwal Please check the edit I did to your question. If there's any mistake please do tell me. — DonAntonio, Feb 14 '18 at 12:22

Jaideep Khare · Accepted Answer · 2018-02-14T13:05:48.493

4

You cannot apply partial limits in addition and subtraction. Though, it can be used in products.

\begin{align} \lim_{x \to 0} \frac{e^x + 1/e^x - 2 \cos x}{x \tan x} &=\lim_{x \to 0} \frac{e^x + 1/e^x - 2 \cos x}{x^2} \times \frac{x}{\tan x }\\ &=\lim_{x \to 0} \frac{e^x + 1/e^x - 2 \cos x}{x^2} \times \lim_{x \to 0} \frac{x}{\tan x }\\ &=\lim_{x \to 0} \frac{e^x \color{red}{-1 -x } + 1/e^x \color{red}{-1 +x +2} - 2 \cos x}{x^2}\\ &=\lim_{x \to 0} \frac{ \color{blue}{e^x -1 -x} + \color{red}{e^{-x} -1 +x} +2 \sin ^2 (x/2)}{x^2}\\ &=\lim_{x \to 0} \frac{ \color{blue}{e^x -1 -x}}{x^2} + \lim_{x \to 0} \frac{\color{red}{e^{-x} -1 +x}}{x^2}+ \lim_{x \to 0} \frac{2 \sin ^2 (x/2)}{x^2}\\ \end{align}

Can you proceed using standard limits now?

edited Feb 14 '18 at 13:05

answered Feb 14 '18 at 12:15

Jaideep Khare

19,293

Yes, I can. Thanks! – Soumil Aggarwal Feb 14 '18 at 12:23
Welcome @Soumil :) – Jaideep Khare Feb 14 '18 at 12:23
I think you want $\frac{x}{\tan x}$ in your first and second lines – prt13463 Feb 14 '18 at 12:32
@prt13463 Yes you're correct. Thanks for noticing. Fixed it. – Jaideep Khare Feb 14 '18 at 13:06

score 1 · Answer 2 · answered Feb 14 '18 at 13:12

Your limit is

$$\lim_{x\to0}\frac{2\cosh x-2\cos x}{x\tan x}=2\lim_{x\to0}\frac{1+2\sinh^2 \dfrac x2-1+2\sin^2 \dfrac x2}{x^2}\frac x{\tan x}=2.$$

(Using $\frac{\sin x}x\to1,\frac{\tan x}x\to1,\frac{\sinh x}x\to1$.)

score 0 · Answer 3 · answered Feb 15 '18 at 15:25

Note that $(\tan x) /x\to 1$ hence the denominator can be replaced by $x^2$. The numerator can be written as $$(e^{x/2}-e^{-x/2})^2+2(1-\cos x) $$ and hence the desired limit is equal to the limit of the expression $$\left(\frac{e^x-1}{x}\right) ^2e^{-x}+2\cdot\frac{1-\cos x} {x^2}$$ and therefore the desired limit is $1^2\cdot 1+2(1/2)=2$.

Your mistake is a very common one. You have replaced the sub-expression $e^x+e^{-x} $ with its limit $2$ while calculating the limit of the given expression. This is invalid and allowed only under certain circumstances.

Unable to evaluate limit correctly

3 Answers3