# What is the limit of (1-7/x)^x as x approaches infinity?

Jun 1, 2016

It is $\frac{1}{e} ^ 7$.

#### Explanation:

We are tempted to substitute and say that $\frac{7}{x}$ goes to zero so it is $1$ to the infinity that is $1$. But this is wrong because it is enough that the quantity in the parenthesis is slightly different from $1$ and the big exponent does the rest.

So it is better to transform it in something that remove the $x$ from the exponent. The best trick is to use the logarithm. So I study the limit of the logarithm and in the end I will do the exponential of the solution.

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

Now the limit is in the form of $\frac{0}{0}$ and I can apply the rule of Hôpital evaluating the derivative:

${\lim}_{x \to \infty} \ln \frac{1 - \frac{7}{x}}{x} ^ - 1$
$= {\lim}_{x \to \infty} \frac{\frac{d}{\mathrm{dx}} \ln \left(1 - \frac{7}{x}\right)}{\frac{d}{\mathrm{dx}} {x}^{-} 1}$

$= {\lim}_{x \to \infty} \frac{\frac{7}{\left(x - 7\right) x}}{- {x}^{-} 2}$

$= {\lim}_{x \to \infty} - \frac{7 {x}^{2}}{\left(x - 7\right) x}$
$= {\lim}_{x \to \infty} - \frac{7 x}{\left(x - 7\right)}$

This form is still $\frac{0}{0}$ so I re-apply Hôpital's rule

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

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

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

This is the limit of the logarithm, so the final result is obtained doing the exponential of this:

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