# Question #3c650

Sep 10, 2017

The limit is zero.

#### Explanation:

As $x \rightarrow {0}^{-} , \frac{1}{x} \rightarrow - \infty .$
For such x, ${e}^{\frac{1}{x}} \rightarrow 0$ and the denominator of the expression goes to 1.
Therefore
As $x \rightarrow {0}^{-} , \frac{x}{1 + {e}^{\frac{1}{x}}} \rightarrow 0.$

As $x \rightarrow {0}^{-} + \frac{1}{x} \rightarrow \infty .$
For such x, ${e}^{\frac{1}{x}} \rightarrow \infty$ and the denominator of the expression goes to $\infty$.
Therefore
As $x \rightarrow {0}^{+} , \frac{1}{1 + {e}^{\frac{1}{x}}} \rightarrow 0 ,$ and
as $x \rightarrow {0}^{+} , \frac{x}{1 + {e}^{\frac{1}{x}}} \rightarrow 0.$

Since the two one-sided limits exist and are equal, the overall limit is zero.