# Question 261f3

Nov 28, 2016

${\lim}_{x \to \infty} {\left(\frac{x + 1}{x + 3}\right)}^{x} = {e}^{-} 2$

#### Explanation:

${\lim}_{x \to \infty} {\left(\frac{x + 1}{x + 3}\right)}^{x} = {\lim}_{x \to \infty} {e}^{\ln} \left({\left(\frac{x + 1}{x + 3}\right)}^{x}\right)$

$= {\lim}_{x \to \infty} {e}^{x \ln \left(\frac{x + 1}{x + 3}\right)}$

=e^(lim_(x->oo)xln((x+1)/(x+3))#

The above step is valid due to the continuity of $f \left(x\right) = {e}^{x}$.

Solving the limit in the exponent, we have

${\lim}_{x \to \infty} x \ln \left(\frac{x + 1}{x + 3}\right) = {\lim}_{x \to \infty} \ln \frac{\frac{x + 1}{x + 3}}{\frac{1}{x}}$

$= {\lim}_{x \to \infty} \frac{\frac{d}{\mathrm{dx}} \ln \left(\frac{x + 1}{x + 3}\right)}{\frac{d}{\mathrm{dx}} \frac{1}{x}}$

The above step applies L'Hopital's rule to a $\frac{0}{0}$ indeterminate form.

$= {\lim}_{x \to \infty} \frac{\frac{d}{\mathrm{dx}} \left(\ln \left(x + 1\right) - \ln \left(x + 3\right)\right)}{- \frac{1}{x} ^ 2}$

$= {\lim}_{x \to \infty} \frac{\frac{1}{x + 1} - \frac{1}{x + 3}}{- \frac{1}{x} ^ 2}$

$= {\lim}_{x \to \infty} \frac{\frac{2}{{x}^{2} + 4 x + 3}}{- \frac{1}{x} ^ 2}$

$= {\lim}_{x \to \infty} - \frac{2 {x}^{2}}{{x}^{2} + 4 x + 3}$

$= {\lim}_{x \to \infty} - \frac{2}{1 + \frac{4}{x} + \frac{3}{x} ^ 2}$

$= - \frac{2}{1 + 0 + 0}$

$= - 2$

Substituting this into our original work, we get the final result:

${\lim}_{x \to \infty} {\left(\frac{x + 1}{x + 3}\right)}^{x} = {e}^{{\lim}_{x \to \infty} x \ln \left(\frac{x + 1}{x + 3}\right)} = {e}^{-} 2$