# How do you find the limit of x Tan(9/x) as x approaches infinity using l'hospital's rule?

May 2, 2016

See below

#### Explanation:

${\lim}_{x \rightarrow \infty} x \tan \left(\frac{9}{x}\right)$ has indeterminate form $\infty \cdot 0$.

In order to use l'Hospital's rule we rewrite the expression:

$x \tan \left(\frac{9}{x}\right) = \frac{\tan \left(\frac{9}{x}\right)}{\frac{1}{x}}$.

The limit is the rewritten expression has form $\frac{0}{0}$, so we can apply l"Hospital.

Differentiating the numerator and denominator gets us:

${\lim}_{x \rightarrow \infty} x \tan \left(\frac{9}{x}\right) = {\lim}_{x \rightarrow \infty} \frac{{\sec}^{2} \left(\frac{9}{x}\right) \left(- \frac{9}{x} ^ 2\right)}{- \frac{1}{x} ^ 2}$

$= {\lim}_{x \rightarrow \infty} \left({\sec}^{2} \left(\frac{9}{x}\right) \left(9\right)\right)$

$= 9 {\sec}^{2} \left(0\right) = 9$

We do not need l'Hospital

$x \tan \left(\frac{9}{x}\right) = 9 \cdot \sin \frac{\frac{9}{x}}{\frac{9}{x}} \cdot \frac{1}{\cos} \left(\frac{9}{x}\right)$

${\lim}_{u \rightarrow \infty} \sin \frac{u}{u} = 1$, so we get

${\lim}_{x \rightarrow \infty} x \tan \left(\frac{9}{x}\right) = {\lim}_{x \rightarrow \infty} 9 \cdot \sin \frac{\frac{9}{x}}{\frac{9}{x}} \cdot \frac{1}{\cos} \left(\frac{9}{x}\right)$

$= \left(9\right) \left(1\right) \left(1\right) = 9$