# What is the limit of lnx-ln(pi)/x-(pi) when x approaches to (pi) from left?

Nov 20, 2016

${\lim}_{x \rightarrow {\pi}^{-}} \frac{\ln \left(x\right) - \ln \left(\pi\right)}{x - \pi} = \frac{1}{\pi}$

#### Explanation:

${\lim}_{x \rightarrow {\pi}^{-}} \frac{\ln \left(x\right) - \ln \left(\pi\right)}{x - \pi}$

Note that this fits the form for the limit definition of the derivative at a point:

$f ' \left(a\right) = {\lim}_{x \rightarrow a} \frac{f \left(x\right) - f \left(a\right)}{x - a}$

So, for ${\lim}_{x \rightarrow {\pi}^{-}} \frac{\ln \left(x\right) - \ln \left(\pi\right)}{x - \pi}$, we see that $f \left(x\right) = \ln \left(x\right)$ and $a = \pi$, so this limit is equivalent to $f ' \left(\pi\right)$ when $f \left(x\right) = \ln \left(x\right)$.

If $f \left(x\right) = \ln \left(x\right)$, then $f ' \left(x\right) = \frac{1}{x}$ and $f ' \left(\pi\right) = \frac{1}{\pi}$. So

$f ' \left(\pi\right) = {\lim}_{x \rightarrow \pi} \frac{f \left(x\right) - f \left(\pi\right)}{x - \pi} = {\lim}_{x \rightarrow {\pi}^{-}} \frac{\ln \left(x\right) - \ln \left(\pi\right)}{x - \pi} = \frac{1}{\pi}$

The sidedness of this limit has no effect on the answer.