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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?

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    $\begingroup$ Hint: Use the taylor expansion of $e^{x-1}$ $\endgroup$ – mrnovice Feb 25 '17 at 0:30
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    $\begingroup$ @RRL Perhaps he cannot take such derivatives yet. However, though, I think this is fine as long as he continues to provide his own efforts and such, as he has done sufficiently well in my opinion. $\endgroup$ – Simply Beautiful Art Feb 25 '17 at 0:48
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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

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

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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,

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


Taylor series solution:

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

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

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  • $\begingroup$ Was expecting some taylor expansion :( $\endgroup$ – bigfocalchord Feb 25 '17 at 0:38
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    $\begingroup$ @dydxx I suppose I can go overkill and wrap all the answers up :P $\endgroup$ – Simply Beautiful Art Feb 25 '17 at 0:39

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