# How do you find the limit of ((16x)/(16x+3))^(4x) as x approaches infinity?

May 17, 2016

The limit is ${e}^{- \frac{3}{4}}$.

#### Explanation:

${\left(\frac{16 x}{16 x + 3}\right)}^{4 x}$,

${\lim}_{x \rightarrow \infty} {\left(\frac{16 x}{16 x + 3}\right)}^{4 x}$ has indeterminate form ${1}^{\infty}$.

We'll try to find the limit of the $\ln$ of the expression and use continuity of the exponential function to get

lim_(xrarroo)((16x)/(16x+3))^(4x) = e^(lim_(xrarroo)ln(((16x)/(16x+3)^)^(4x))

$\ln \left({\left(\frac{16 x}{16 x + 3}\right)}^{4 x}\right) = 4 x \ln \left(\frac{16 x}{16 x + 3}\right)$

$= 4 \ln \frac{\frac{16 x}{16 x + 3}}{\frac{1}{x}}$

As $x \rightarrow \infty$, this has indeterminate form $\frac{0}{0}$, so we can use l"Hospital's rule.

${\lim}_{x \rightarrow \infty} 4 \ln \frac{\frac{16 x}{16 x + 3}}{\frac{1}{x}} = 4 {\lim}_{x \rightarrow \infty} \frac{\frac{16 x + 3}{16 x} \left(\frac{16 \left(16 x + 3\right) - \left(16 x\right) \left(16\right)}{16 x + 3} ^ 2\right)}{- \frac{1}{x} ^ 2}$

$= 4 {\lim}_{x \rightarrow \infty} \frac{\frac{48}{16 x \left(16 x + 3\right)}}{- \frac{1}{x} ^ 2}$

 = 4lim_(xrarroo)(-48x^2)/(16x(16x+3)))

$= 4 \left(\frac{- 48}{16} ^ 2\right) = 4 \left(\frac{- 3}{16}\right) = - \frac{3}{4}$

So

${\lim}_{x \rightarrow \infty} {\left(\frac{16 x}{16 x + 3}\right)}^{4 x} = {e}^{- \frac{3}{4}}$

May 17, 2016

This limit can be evaluated without resorting to l'Hospital's rule using ${\lim}_{u \rightarrow \infty} {\left(1 + \frac{1}{u}\right)}^{u} = e$

#### Explanation:

${\left(\frac{16 x}{16 x + 3}\right)}^{4 x} = {\left(\frac{\left(16 x + 3\right) - 3}{16 x + 3}\right)}^{4 x}$

$= {\left(1 + \frac{- 3}{16 x + 3}\right)}^{4 x}$

$= {\left(1 + \frac{1}{\frac{- \left(16 x + 3\right)}{3}}\right)}^{4 x}$

$= {\left(1 + \frac{1}{\frac{- \left(16 x + 3\right)}{3}}\right)}^{4 x}$

We need to make the exponent equal to $\frac{- \left(16 x + 3\right)}{3}$ so we'll multiply the exponent by $\frac{- \left(16 x + 3\right)}{12 x}$ and also by the reciprocal.

$= {\left({\left(1 + \frac{1}{\frac{- \left(16 x + 3\right)}{3}}\right)}^{\left(4 x\right) \frac{- \left(16 x + 3\right)}{12 x}}\right)}^{\frac{- 12 x}{16 x + 3}}$

$= {\left({\left(1 + \frac{1}{\frac{- \left(16 x + 3\right)}{3}}\right)}^{\frac{- \left(16 x + 3\right)}{3}}\right)}^{\frac{- 12 x}{16 x + 3}}$

As $x \rightarrow \infty$, we have $\left(\frac{- \left(16 x + 3\right)}{3}\right) \rightarrow \infty$ and $\frac{- 12 x}{16 x + 3} \rightarrow - \frac{3}{4}$

Consequently,

${\lim}_{x \rightarrow \infty} {\left(\frac{16 x}{16 x + 3}\right)}^{4 x} = {\lim}_{x \rightarrow \infty} {\left[{\left(1 + \frac{1}{\frac{- \left(16 x + 3\right)}{3}}\right)}^{\frac{- \left(16 x + 3\right)}{3}}\right]}^{\frac{- 12 x}{16 x + 3}}$

$= {\left[e\right]}^{- \frac{3}{4}}$