Unable to evaluate limit correctly 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.
 A: 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?
A: 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 $\frac{\sin x}x\to1,\frac{\tan x}x\to1,\frac{\sinh x}x\to1$.)
A: 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. 
