Find $\lim\limits_{n\to\infty}{e^n - e^{\frac1n + n}}$ So far I've tried: $${e^n - e^{\frac1n + n}} = e^n(1- e^{\frac1n}).$$ Then appling l'Hopitals rule to $$\lim\limits_{n\to\infty} e^n(1- e^{\frac1n}) = \lim\limits_{n\to\infty} \dfrac{(1- e^{\frac1n})}{e^{-n}},$$ I have not found success. Is there another way to manipulate the expression to be able to apply l'Hopitals? I know the limit should approach -$\infty$. To clarify, I can use l'Hopital.
 A: HINT
We have that
$${e^n - e^{\frac1n + n}}=e^n \left(1-e^{\frac1n}\right)=-\frac{e^n}n \frac{e^{\frac1n}-1}{\frac1n}$$
then use standard limits.
A: The answer is $-\infty$.
The trick is to factor out an $e^{n}$ term, and force L'Hopital's Rule by writing the expression as a fraction. We have
$$\lim_{n\to\infty} e^{n} - e^{\frac{1}{n} + n} = \lim_{n\to\infty} e^{n}\left(1 - e^{1/n}\right)$$
$$= \lim_{n\to\infty} \frac{e^{n}\left(1 - e^{1/n}\right)\left(1 + e^{1/n}\right)}{(1 + e^{1/n})} $$
$$= \lim_{n\to\infty}\frac{e^{n} \left(1 - e^{2/n}\right)}{1 + e^{1/n}} $$
$$= \lim_{n\to\infty} \frac{e^{n} - e^{n + \frac{2}{n}}}{1 + e^{1/n}}.$$
By L'Hopital's Rule, the above expression equals
$$\lim_{n\to\infty} \frac{e^{n} - e^{n + \frac{2}{n}}\left(1 - \frac{2}{n^{2}}\right)}{-\frac{1}{n^{2}} \cdot e^{1/n}}  \\[1em] $$
$$= \lim_{n\to\infty} \frac{-n^{2} \left( e^{n} - e^{n + \frac{2}{n}}\left(1 - \frac{2}{n}\right)\right)}{e^{1/n}}.$$
As $n \rightarrow \infty$, $e^{1/n}$ approaches $1$, and the numerator clearly approaches $-\infty$. 
Therefore, the answer is $-\infty$.
