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.

Answer 1

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?

Answer 2

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

Answer 3

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.