# Prove that lim_(x rarr2) ( 2^x-4 ) / (x-2) =ln16?

Jan 30, 2018

$f ' \left(2\right) = \ln 16 \setminus \setminus \setminus$ QED

#### Explanation:

Given the context of the question, we can see that this limit is the derivative of the function $f \left(x\right) = {2}^{x}$ when $x = 2$, Using the limit definition of the derivative, we calculate this using:

$f ' \left(2\right) = {\lim}_{x \rightarrow 2} \frac{f \left(x\right) - f \left(2\right)}{x - 2}$
$\setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus = {\lim}_{x \rightarrow 2} \frac{{2}^{x} - {2}^{2}}{x - 2}$
$\setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus = {\lim}_{x \rightarrow 2} \frac{{2}^{x} - 4}{x - 2}$

Let us perform a substitution, where:

$u = x - 2$

Clearly as $x \rightarrow 2 \implies u \rightarrow 0$, and so we can rewrite the limits as:

$f ' \left(2\right) = {\lim}_{u \rightarrow 0} \frac{{2}^{u + 2} - 4}{u}$
$\setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus = {\lim}_{u \rightarrow 0} \frac{{2}^{2} {2}^{u} - 4}{u}$
$\setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus = 4 \setminus {\lim}_{u \rightarrow 0} \frac{{2}^{u} - 1}{u}$

Now we require some additional knowledge of logarithms and specific calculus limits, which provides the value of the limit

${\lim}_{u \rightarrow 0} \frac{{a}^{u} - 1}{u} = \ln a$

Where $\ln$ denotes the Natural Logarithm, ie a logarithm to base $e$, where $e$ is Euler's Number. Using this result we get:

$f ' \left(2\right) = 4 \setminus \ln 2$
$\setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus = \ln {2}^{4}$
$\setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus = \ln 16 \setminus \setminus \setminus$ QED