Evaluate $\lim_{x\to1}\frac{e^x-ex}{(x-1)^2}$ using Taylor's series

Question

I am trying to evaluate

$$\lim_{x\to1}\frac{e^x-ex}{(x-1)^2}$$

using the Taylor series.

I am thinking I can use the Taylor series of $e^x$ at $x=1$ for this which I found to be $$e+e(x-1)+\frac{e}{2}(x-1)^2.$$ However, I don't know how to maneuver the expression $\frac{e^x-ex}{(x-1)^2}$ in a way that I can substitute part of it with the Taylor's series.

Should I try another series or how should I continue?

Answer 1

$$ \begin{split} \lim_{x \to 1} \frac{\operatorname{e}^x-\operatorname{e}x}{(x-1)^2} &= \lim_{x \to 1} \frac{\operatorname{e}+\operatorname{e}(x-1)+\frac{\operatorname{e}}{2}(x-1)^2+o(x-1)^3-\operatorname{e}x}{(x-1)^2}\\ &=\lim_{x \to 1} \frac{\operatorname{e}+\operatorname{e}x-\operatorname{e}+\frac{\operatorname{e}}{2}(x-1)^2+o(x-1)^3-\operatorname{e}x}{(x-1)^2}\\ &=\lim_{x \to 1} \frac{\frac{\operatorname{e}}{2}(x-1)^2+o(x-1)^3}{(x-1)^2}\\ &= \lim_{x \to 1}\left[\frac{\operatorname{e}}{2}+ \frac{o(x-1)^3}{(x-1)^2}\right]\\ &= \lim_{x \to 1}\left[\frac{\operatorname{e}}{2}+ o(x-1)\right]\\ &= \frac{\operatorname{e}}{2}+\lim_{x \to 1} o(x-1) = \frac{\operatorname{e}}{2}+ 0 = \frac{\operatorname{e}}{2}\\ \end{split} \\ $$

Answer 2

You can simply use the Taylor series for $e^x$ in the numerator of your expression, and note that the terms proportional to $(x-1)^0$ and $(x-1)^1$ cancel out.

Answer 3

$$\lim_{x \to1}\frac{e^x-ex}{(x-1)^2}$$ $$=\lim_{x \to1}\frac{ee^{x-1}-e(x-1)-e}{(x-1)^2}$$ $$=\lim_{y \to 0}\frac{ee^y-ey-e}{y^2}$$ $$=\lim_{y \to 0}\frac{e(1+y+\frac{y^2}{2!}+\frac{y^3}{3!}+...)-ey-e}{y^2}$$ $$=\lim_{y \to 0}\frac{e(\frac{y^2}{2!}+\frac{y^3}{3!}+...)}{y^2}$$ $$=\lim_{y \to 0}e(\frac{1}{2!}+\frac{y}{3!}+...)$$ $$=\frac{e}{2}.$$