# Question #58b59

Jan 10, 2018

$27 \ln \left(3\right) + 1$

#### Explanation:

Direct substitution gives $\frac{0}{0}$ so we can use L'Hopital's rule:

${\lim}_{x \to 3} \frac{{3}^{x} - 27 + \ln \left(x - 2\right)}{x - 3} =$
${\lim}_{x \to 3} \frac{\frac{d}{\mathrm{dx}} \left({3}^{x} - 27 + \ln \left(x - 2\right)\right)}{\frac{d}{\mathrm{dx}} \left(x - 3\right)} =$
${\lim}_{x \to 3} \frac{{3}^{x} \cdot \ln \left(3\right) + \left(\frac{1}{x - 2}\right)}{1}$

now we can substitute:

${\lim}_{x \to 3} \frac{{3}^{x} \cdot \ln \left(3\right) + \left(\frac{1}{x - 2}\right)}{1} = 27 \ln \left(3\right) + 1$

A second approach using just derivatives:

If you take the original limit and break it into two limits you have:

${\lim}_{x \to 3} \frac{{3}^{x} - 27 + \ln \left(x - 2\right)}{x - 3} =$
${\lim}_{x \to 3} \frac{{3}^{x} - 27}{x - 3} + {\lim}_{x \to 3} \frac{\ln \left(x - 2\right)}{x - 3}$

notice that $\ln \left(3 - 2\right) = \ln \left(1\right) = 0$ so we can add one more detail to our new limits to get:

${\lim}_{x \to 3} \frac{{3}^{x} - 27}{x - 3} + {\lim}_{x \to 3} \frac{\ln \left(x - 2\right) - \ln \left(3 - 2\right)}{x - 3}$

Let's look at the left limit:
${\lim}_{x \to 3} \frac{{3}^{x} - 27}{x - 3}$

This is the definition of the derivative of the function $f \left(x\right) = {3}^{x}$ at $x = 3$.

For $f \left(x\right) = {3}^{x}$ we know $f ' \left(x\right) = {3}^{x} \cdot \ln \left(3\right)$, so:

${\lim}_{x \to 3} \frac{{3}^{x} - 27}{x - 3} = f ' \left(3\right) = {3}^{3} \ln \left(3\right) = 27 \ln \left(3\right)$

For the right limit, it's the definition of the derivative of $f \left(x\right) = \ln \left(x - 2\right)$ at $x = 3$.

For $f \left(x\right) = \ln \left(x - 2\right)$ we know that $f ' \left(x\right) = \frac{1}{x - 2}$, so:

${\lim}_{x \to 3} \frac{\ln \left(x - 2\right) - \ln \left(3 - 2\right)}{x - 3} = f ' \left(3\right) = \frac{1}{3 - 2} = 1$

So the overall limit will be the sum of the values of these two derivatives:

${\lim}_{x \to 3} \frac{{3}^{x} - 27}{x - 3} + {\lim}_{x \to 3} \frac{\ln \left(x - 2\right) - \ln \left(3 - 2\right)}{x - 3} = 27 \ln \left(3\right) + 1$