Need help with this limit $\lim \limits_{x\to \:1}\left(\frac{xe^{x^2}-e}{x-1}\right)$ I haven't been able to solve this limit, so help would be appreciated.
L'Hopital is not allowed.
Thanks in advance.
Hint:

Putting $f(x) = x e^{x^2}$, that limit is just $f'(1)$.
\begin{align}
L &amp;= \lim_{x \to 1}\frac{xe^{x^{2}} - e}{x - 1}\notag\\
&amp;= \lim_{x \to 1}\frac{xe^{x^{2}} - e}{x^{2} - 1}\cdot\frac{x^{2} - 1}{x - 1}\notag\\
&amp;= \lim_{x \to 1}\frac{xe^{x^{2}} - e}{x^{2} - 1}\cdot(x + 1)\notag\\
&amp;= 2\lim_{x \to 1}\frac{xe^{x^{2}} - xe + xe - e}{x^{2} - 1}\notag\\
&amp;= 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\\
&amp;= 2\lim_{x \to 1}\left(xe\cdot\frac{e^{x^{2} - 1} - 1}{x^{2} - 1} + \frac{e}{x + 1}\right)\notag\\
&amp;= e + 2e\lim_{x \to 1}\frac{e^{x^{2} - 1} - 1}{x^{2} - 1}\notag\\
&amp;= e + 2e\lim_{t \to 0}\frac{e^{t} - 1}{t}\text{ (by putting }t = x^{2} - 1)\notag\\
&amp;= e + 2e = 3e\notag
\end{align}