# How do you find the limit of x^(7/x) as x approaches infinity?

Mar 23, 2016

${\lim}_{x \to \infty} {x}^{\frac{7}{x}} = 1$

#### Explanation:

First, we will use the following:

• ${e}^{\ln} \left(x\right) = x$
• Because ${e}^{x}$ is continuous on $\left(- \infty , \infty\right)$, we have ${\lim}_{x \to \infty} {e}^{f} \left(x\right) = {e}^{{\lim}_{x \to \infty} f \left(x\right)}$

With these:

${\lim}_{x \to \infty} {x}^{\frac{7}{x}} = {\lim}_{x \to \infty} {e}^{\ln \left({x}^{\frac{7}{x}}\right)}$

$= {\lim}_{x \to \infty} {e}^{\frac{7}{x} \ln \left(x\right)}$

$= {e}^{{\lim}_{x \to \infty} \frac{7}{x} \ln \left(x\right)}$

Next, we will use L'Hopital's rule:

${\lim}_{x \to \infty} \frac{7}{x} \ln \left(x\right) = {\lim}_{x \to \infty} \frac{7 \ln \left(x\right)}{x}$

$= {\lim}_{x \to \infty} \frac{\frac{d}{\mathrm{dx}} 7 \ln \left(x\right)}{\frac{d}{\mathrm{dx}} x}$

$= {\lim}_{x \to \infty} \frac{\frac{7}{x}}{1}$

$= {\lim}_{x \to \infty} \frac{7}{x}$

$= 0$

Putting these together, we get our final result:

${\lim}_{x \to \infty} {x}^{\frac{7}{x}} = {e}^{{\lim}_{x \to \infty} \frac{7}{x} \ln \left(x\right)} = {e}^{0} = 1$