# How do you find the limit of y=lnx/cscx  as x approaches 0 using l'hospital's rule?

Apr 28, 2016

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

#### Explanation:

First, assuming we are only consider real numbers, we can only take the limit of $\ln \left(x\right)$ as $x$ approaches $0$ from the positive direction, and so we will proceed as such.

Noting that ${\lim}_{x \to {0}^{+}} \ln \left(x\right) = - \infty$ and ${\lim}_{x \to {0}^{+}} \csc \left(x\right) = \infty$, we have ${\lim}_{x \to {0}^{+}} \ln \frac{x}{\csc} \left(x\right)$ as a $\frac{\infty}{\infty}$ indeterminate form. As such, we can apply L'hopital's rule.

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

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

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

If two functions converge at a point, then their product also converges at that point, and the limit of the product is the product of the limits of the original functions. Using that fact, together with the known limits ${\lim}_{x \to 0} \sin \frac{x}{x} = 1$, and ${\lim}_{x \to 0} \tan \left(x\right) = 0$, we can factor our expression into the product of two limits:

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

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

$= - 1 \cdot 0$

$= 0$