# How do you evaluate [ (ln x) / (csc x) ] as x approaches 0+?

Jan 17, 2017

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

#### Explanation:

You can write the function as:

$\ln \frac{x}{\csc} x = \ln x \sin x = \left(x \ln x\right) \left(\sin \frac{x}{x}\right)$

so that:

${\lim}_{x \to {0}^{+}} \ln \frac{x}{\csc} x = \left({\lim}_{x \to {0}^{+}} \left(x \ln x\right)\right) \cdot \left({\lim}_{x \to {0}^{+}} \left(\sin \frac{x}{x}\right)\right)$

Both limits are well known, but we can remind how they can be calculated:

${\lim}_{x \to {0}^{+}} \left(\sin \frac{x}{x}\right)$

is in the indeterminate form $\frac{0}{0}$ and can be calculated using l'Hospital's rule:

${\lim}_{x \to {0}^{+}} \left(\sin \frac{x}{x}\right) = {\lim}_{x \to {0}^{+}} \frac{\frac{d}{\mathrm{dx}} \sin x}{\frac{d}{\mathrm{dx}} x} = {\lim}_{x \to {0}^{+}} \cos \frac{x}{1} = 1$

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

is in the indeterminate form $\frac{0}{\infty}$ but can be reduced to a different form by expressing it as a quotient:

${\lim}_{x \to {0}^{+}} \left(x \ln x\right) = {\lim}_{x \to {0}^{+}} \ln \frac{x}{\frac{1}{x}}$

This is now in the form $\frac{\infty}{\infty}$ and we apply again l'Hospital's rule:

${\lim}_{x \to {0}^{+}} \ln \frac{x}{\frac{1}{x}} = {\lim}_{x \to {0}^{+}} \frac{\frac{d}{\mathrm{dx}} \ln x}{\frac{d}{\mathrm{dx}} \left(\frac{1}{x}\right)} = {\lim}_{x \to {0}^{+}} \frac{\frac{1}{x}}{- \frac{1}{x} ^ 2} = {\lim}_{x \to {0}^{+}} \left(- x\right) = 0$

We can then conclude that:

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