# How do you evaluate lim_(x->0^(-)) e^x/(1+lnx)?

Dec 28, 2016

Since the limit of the denominator does not readily tend to $0$, we can use the quotient rule of limits.

However, it makes more sense to evaluate the right-handed limit, since to the left of $x = 0$, the function doesn't exist in the reals.

$\textcolor{b l u e}{{\lim}_{x \to {0}^{-}} {e}^{x} / \left(1 + \ln x\right)}$

$\implies {\lim}_{x \to {0}^{+}} {e}^{x} / \left(1 + \ln x\right)$

(where we have switched to the right-handed limit because the function is clearly continuous to the right of $x = 0$, and at $x = 0$ as well.)

$= \frac{{\lim}_{x \to {0}^{+}} {e}^{x}}{{\lim}_{x \to {0}^{+}} 1 + \ln x}$

$= \frac{1}{- \infty}$

$= \textcolor{b l u e}{0}$

This is shown below:

graph{e^x/(1 + lnx) [-1, 4, -10.78, 14.88]}

Although there is an existent limit at x = 0, it, again, doesn't make as much sense to think about how it's evaluated from the left, as it does to consider the righthand limit.

Dec 28, 2016

$f \left(x\right) = {e}^{x} / \left(1 + \ln x\right)$ is not defined for $x < 0$, so:

${\lim}_{x \to {0}^{-}} f \left(x\right)$ has no meaning.