How to deal with division of two summations

Question

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.

Answer 1

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

$=2\lim\limits_{x\to0}\dfrac{x+O\left(x^3\right)}{x-O\left(x^3\right)}=$

$=2\lim\limits_{x\to0}\dfrac{x\left(1+O\left(x^2\right)\right)}{x\left(1-O\left(x^2\right)\right)}=$

$=2\lim\limits_{x\to0}\dfrac{1+O\left(x^2\right)}{1-O\left(x^2\right)}=$

$=2\cdot1=$

$=2\;.$

Another way is to use notable limits.

$\lim\limits_{x\to 0}\dfrac{e^x-e^{-x}}{\sin(x)}=$

$=\lim\limits_{x\to0}\left[\dfrac x{\sin x}\left(\dfrac{e^x-1}x+\dfrac{e^{-x}-1}{-x}\right)\right]=$

$=\lim\limits_{x\to0}\dfrac x{\sin x}\cdot\left(\lim\limits_{x\to0}\dfrac{e^x-1}x+\lim\limits_{x\to0}\dfrac{e^{-x}-1}{-x}\right)=$

$=1\cdot(1+1)=$

$=2\;.$

Answer 2

By Taylor to the first order, $$\frac{2\sinh(x)}{\sin(x)}=\frac{2x+o(x)}{x+o(x)}\to2.$$

Answer 3

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*}

Answer 4

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)