How to deal with division of two summations Actually, I was solving this question.

$$\displaystyle\lim_{x\to
 0}\frac{e^x-e^{-x}}{\sin(x)}=\displaystyle\lim_{x\to
 0}\frac{2\sinh(x)}{\sin(x)}=2\displaystyle\lim_{x\to
 0}\left(\frac{\displaystyle\sum_{n=0}^\infty
\frac{x^{2n+1}}{(2n+1)!}}{\displaystyle\sum_{n=0}^\infty\frac{(-1)^n}{(2n+1)!}x^{2n+1}}\right)$$

How to now deal with these two summations? Or is my way itself wrong?
Any hint would be appreciated.
 A: $2\displaystyle\lim_{x\to
 0}\left(\frac{\displaystyle\sum_{n=0}^\infty
\frac{x^{2n+1}}{(2n+1)!}}{\displaystyle\sum_{n=0}^\infty\frac{(-1)^n}{(2n+1)!}x^{2n+1}}\right)=$
$=2\lim\limits_{x\to0}\dfrac{x+O\left(x^3\right)}{x-O\left(x^3\right)}=$
$=2\lim\limits_{x\to0}\dfrac{x\left(1+O\left(x^2\right)\right)}{x\left(1-O\left(x^2\right)\right)}=$
$=2\lim\limits_{x\to0}\dfrac{1+O\left(x^2\right)}{1-O\left(x^2\right)}=$
$=2\cdot1=$
$=2\;.$
Another way is to use notable limits.
$\lim\limits_{x\to
 0}\dfrac{e^x-e^{-x}}{\sin(x)}=$
$=\lim\limits_{x\to0}\left[\dfrac x{\sin x}\left(\dfrac{e^x-1}x+\dfrac{e^{-x}-1}{-x}\right)\right]=$
$=\lim\limits_{x\to0}\dfrac x{\sin x}\cdot\left(\lim\limits_{x\to0}\dfrac{e^x-1}x+\lim\limits_{x\to0}\dfrac{e^{-x}-1}{-x}\right)=$
$=1\cdot(1+1)=$
$=2\;.$
A: By Taylor to the first order, $$\frac{2\sinh(x)}{\sin(x)}=\frac{2x+o(x)}{x+o(x)}\to2.$$
A: We can take the summation of $\csc(x)$ instead of $\sin(x)$ in the denominator. For instance,
\begin{align*}
\csc x &= \sum_{n \mathop = 0}^\infty \dfrac { (-1)^{n - 1} 2 ({2^{2 n - 1} - 1}) B_{2 n} \, x^{2 n - 1} } {(2 n)!}\\
&= \frac 1 x + \frac x 6 + \frac {7 x^3} {360} + \frac {31 x^5} {15 \, 120} + \cdots
\end{align*}
Similarly, $$\sinh(x) = x + \frac{x^3}{3!} + \dots$$
Taking the product of the above two series, under the limit $x\to 0$ all other terms go to 0 and only the constant term survives, which is equal to 1.
Therefore $\lim\limits_{x\to 0}\frac{2\sinh(x)}{\sin(x)} = 2$
(which can also be obtained using the L'hopital rule)
A: There is no need to evaluate the entire summation, just consider the first two or three terms should give us the reuired limit.
\begin{gather*}
\sum ^{\infty }_{i=0}\frac{x^{2i+1}}{( 2i+1) !} =x+\frac{x^{3}}{3!} +\frac{x^{5}}{5!} +O\left( x^{7}\right)\\
\sum ^{\infty }_{i=0}\frac{( -1)^{i} x^{2i+1}}{( 2i+1) !} =x-\frac{x^{3}}{3!} +\frac{x^{5}}{5!} +O\left( x^{7}\right)\\
Hence,\\
\lim _{x\rightarrow 0}\frac{\sum ^{\infty }_{i=0}\frac{x^{2i+1}}{( 2i+1) !}}{\sum ^{\infty }_{i=0}\frac{( -1)^{i} x^{2i+1}}{( 2i+1) !}} =\frac{x+\frac{x^{3}}{3!} +\frac{x^{5}}{5!} +O\left( x^{7}\right)}{x-\frac{x^{3}}{3!} +\frac{x^{5}}{5!} +O\left( x^{7}\right)} =1
\end{gather*}
