Solve $\lim_{x \rightarrow 2} \frac{xe^{x-1}-2e}{x-2}$ without using L'Hopital's rule

Question

I tried:

$$\lim_{x \rightarrow 2} \frac{xe^{x-1}-2e}{x-2} = \frac{(x-2)?}{x-2}$$

I feel like I can factor $xe^{x-1}-2e$ into $(x-2)y$, but I am not sure what y is.

I tried figuring it out and the closes I got was $$(x-2)(xe^{x-1}e) = xe^{x-1}+xe-2e^{x-1}-2e$$

Am I on the right track? How do I solve this?

Answer 1

Simply use the definition of derivative. Let $f(x)=xe^{x-1}$, then $f(2)=2e$.

Also, $f'(x)=(x+1)e^{x-1}$

So the limit becomes

$\lim_{x \to 2} \frac {f(x)-f(2)}{x-2} =f'(2)=3e$

Answer 2

Algebra solution:

Let $x=u+2$.

$$\begin{align}\frac{xe^{x-1}-2e}{x-2}&=\frac{xe^{x-1}-2e^{x-1}+2e^{x-1}-2e}{x-2}\\&=\frac{(x-2)e^{x-1}+2e(e^{x-2}-1)}{x-2}\\&=e^{x-1}+2e\frac{e^u-1}u\end{align}$$

We know that $\frac{e^u-1}u\to1$, thus,

$$\lim_{x\to2}\frac{xe^{x-1}-2e^{2-1}}{x-2}=e+2e=3e$$

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Taylor series solution:

Let $x=u+2$ and $e^u=1+u+\mathcal O(u^2)$.

$$\begin{align}\frac{xe^{x-1}-2e}{x-2}&=e\frac{(u+2)e^u-2}u\\&=e\frac{(u+2)(1+u+\mathcal O(u^2))-2}u\\&=e\frac{3u+\mathcal O(u^2)}u\\&=3e+\mathcal O(u)\\&\to3e\end{align}$$