Jtdylktuy

So far I've tried: $${e^n - e^{frac1n + n}} = e^n(1- e^{frac1n}).$$ Then appling l'Hopitals rule to $$limlimits_{ntoinfty} e^n(1- e^{frac1n}) = limlimits_{ntoinfty} dfrac{(1- e^{frac1n})}{e^{-n}},$$ 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.

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.

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

$$lim_{ntoinfty} e^{n} - e^{frac{1}{n} + n} = lim_{ntoinfty} e^{n}left(1 - e^{1/n}right)$$

$$= lim_{ntoinfty} frac{e^{n}left(1 - e^{1/n}right)left(1 + e^{1/n}right)}{(1 + e^{1/n})} $$

$$= lim_{ntoinfty}frac{e^{n} left(1 - e^{2/n}right)}{1 + e^{1/n}} $$

$$= lim_{ntoinfty} frac{e^{n} - e^{n + frac{2}{n}}}{1 + e^{1/n}}.$$

By L'Hopital's Rule, the above expression equals

$$lim_{ntoinfty} frac{e^{n} - e^{n + frac{2}{n}}left(1 - frac{2}{n^{2}}right)}{-frac{1}{n^{2}} cdot e^{1/n}} \[1em] $$

$$= lim_{ntoinfty} frac{-n^{2} left( e^{n} - e^{n + frac{2}{n}}left(1 - frac{2}{n}right)right)}{e^{1/n}}.$$

As $n rightarrow infty$, $e^{1/n}$ approaches $1$, and the numerator clearly approaches $-infty$.

Therefore, the answer is $-infty$.