# Lim x-> infinity 7/((e^x)+2) evaluate ?

Jan 19, 2018

${\lim}_{x \to \infty} \frac{7}{{e}^{x} + 2} = 0$

#### Explanation:

Let try to substitute $\infty$ for $x$

Recall: ${\lim}_{x \to \infty} {e}^{x} = \infty$

This can easily be verified by taking a look of the graph of ${e}^{x}$ (See Below)

graph{e^x [-9.58, 10.42, -1.24, 8.76]}

${\lim}_{x \to \infty} \frac{7}{{e}^{x} + 2} = \frac{7}{{e}^{\textcolor{red}{\infty}} + 2} = \frac{7}{\infty + 2} = \frac{7}{\infty} = 0$

As to why $\frac{7}{\infty}$ tends to $0$ think of it like this:

As $x$ approaches infinity the denominator gets larger and larger towards infinity but the fraction as a whole gets smaller and smaller. Here's a few examples:

Consider:

$\frac{1}{10} = 0.1$

$\frac{1}{100} = 0.01$

$\frac{1}{1000} = 0.001$

$\frac{1}{10000000} = 0.0000001$

As you can see, the fraction becomes smaller and smaller/ closer and closer to $0$. The fact that the numerator is $7$ instead of $1$ doesn't change this fact.