Find $lim_{x \to \pi} {\frac{\sin(x)}{e^{\pi}-e^x}}$ using series expansion

Question

Using L'Hospital's Rule ($\frac{0}{0}$ Case): $$\lim_{x \to \pi} {\frac{\sin(x)}{e^{\pi}-e^x}}=\lim_{x \to \pi} {\frac{\cos(x)}{-e^x}}=\frac{-1}{-e^{\pi}}=\frac{1}{e^{\pi}}$$

But can I also do this using series expansions. For example:

$$\lim_{x \to \pi} \frac{x-\frac{x^3}{3!}+...}{e^{\pi}-[e^{\pi}(x-\pi)+e^{\pi}\frac{1}{2}(x-\pi)^2+...]}$$

Which then simplifies to: $$\lim_{x \to \pi} \frac{x-\frac{x^3}{3!}+...}{-e^{\pi}[(x-\pi)+\frac{1}{2}(x-\pi)^2+...]}$$

And now I don't know how to proceed.

Answer 1

Use the fact that $\sin x=\sin (\pi -x)$. You have $\frac{(\pi -x)-\frac{(\pi -x)^3}{3!}+...}{-e^{\pi}[(x-\pi)+\frac{1}{2}(x-\pi)^2+...]}$. Cancel one factor of $x-\pi$ from numerator and denominator and put $x=\pi$.