Need help with this limit $\lim \limits_{x\to \:1}\left(\frac{xe^{x^2}-e}{x-1}\right)$

Question

I haven't been able to solve this limit, so help would be appreciated. L'Hopital is not allowed. Thanks in advance.

Answer 1

HINT:

Set $x-1=h$

$$\lim_{h\to0}\dfrac{(h+1)e^{(1+h)^2}-e}h=\lim_{h\to0}e^{(1+h)^2}+e\cdot\lim_{h\to0}\dfrac{e^{2h+h^2}-1}h$$

Now $$\lim_{h\to0}\dfrac{e^{2h+h^2}-1}{2h+h^2}\cdot\lim_{h\to0}\dfrac{2h+h^2}h$$

Answer 2

Hint:

Putting $f(x) = x e^{x^2}$, that limit is just $f'(1)$.

Answer 3

Let $f(x) = xe^{x^2}.$ The limit we want is

$$\lim_{x\to 1}\frac{f(x)- f(1)}{x-1}.$$

By definition, this is $f'(1).$ So just calculate $f'(1)$ (and no, we're not using L'Hopital here).

Answer 4

The best way to proceed has already been presented. I thought that it might be instructive to see another approach. To that end, we write

$$e^{x^2}=e+e(x^2-1)+O(x^2-1)^2$$

so that

$$\begin{align} \frac{xe^{x^2}-e}{x-1}&=\frac{e(x-1)+e(x^2-1)^2+O(x^2-1)^2}{x-1}\\\\ &=3e+O(x-1)\\\\ &\to 3e\,\,\text{as}\,\,x\to 1 \end{align}$$

Answer 5

\begin{align} L &= \lim_{x \to 1}\frac{xe^{x^{2}} - e}{x - 1}\notag\\ &= \lim_{x \to 1}\frac{xe^{x^{2}} - e}{x^{2} - 1}\cdot\frac{x^{2} - 1}{x - 1}\notag\\ &= \lim_{x \to 1}\frac{xe^{x^{2}} - e}{x^{2} - 1}\cdot(x + 1)\notag\\ &= 2\lim_{x \to 1}\frac{xe^{x^{2}} - xe + xe - e}{x^{2} - 1}\notag\\ &= 2\lim_{x \to 1}xe\cdot\frac{e^{x^{2} - 1} - 1}{x^{2} - 1} + e\cdot\frac{x - 1}{x^{2} - 1}\notag\\ &= 2\lim_{x \to 1}\left(xe\cdot\frac{e^{x^{2} - 1} - 1}{x^{2} - 1} + \frac{e}{x + 1}\right)\notag\\ &= e + 2e\lim_{x \to 1}\frac{e^{x^{2} - 1} - 1}{x^{2} - 1}\notag\\ &= e + 2e\lim_{t \to 0}\frac{e^{t} - 1}{t}\text{ (by putting }t = x^{2} - 1)\notag\\ &= e + 2e = 3e\notag \end{align}