# l'Hôpital's Rule

• November 18th 2007, 07:04 AM
siriano.3
l'Hôpital's Rule
lim [ln(x^4-3)]/[ln(x)cos(1/x)]
x-> INF

So, that is the form INF/INF, so I tried to take the derivative of the numerator and denominator, but I can't figure out the derivative of the denominator.
• November 18th 2007, 07:07 AM
Jhevon
Quote:

Originally Posted by siriano.3
lim [ln(x^4-3)]/[ln(x)cos(1/x)]
x-> INF

So, that is the form INF/INF, so I tried to take the derivative of the numerator and denominator, but I can't figure out the derivative of the denominator.

it's a doozie!

by the product rule and chain rule:

$\frac d{dx} \ln x \cos \left( \frac 1x \right) = \frac 1x \cos \left( \frac 1x \right) + \frac {\ln x}{x^2} \sin \left( \frac 1x \right)$

try seeing what you can simplify before applying L'Hopital's
• November 18th 2007, 09:32 AM
ThePerfectHacker
$\lim_{x\to \infty} \frac{\ln (x^4-3)}{\ln x \cos \frac{1}{x}}$ since $\lim_{x\to \infty} \frac{1}{\cos \frac{1}{x}} = 1$. This problem reduces to:
$\lim_{x\to \infty} \frac{\ln (x^4-3)}{\ln x}$.
We can solve this without L'Hopital, note when $x$ is large:
$\ln (x^4) - \ln (3) \leq \ln (x^4 -3) \leq \ln (x^4)$
Thus,
$\frac{\ln x^4 - \ln 3}{\ln x} \leq \frac{\ln (x^4-3)}{\ln x} \leq \frac{\ln x^4}{\ln x}$.
Now the left side and right side limits are 4, because, for example,
$\lim_{x\to \infty} \frac{\ln x^4 - \ln 3}{\ln x} = \lim_{x\to \infty} \frac{\ln x^4}{\ln x} - \underbrace{\frac{\ln 3}{\ln x}}_{\mbox{ this is zero }}$.
But,
$\lim_{x\to \infty} \frac{\ln x^4}{\ln x} = \lim_{x\to \infty} \frac{4\ln x}{\ln x} = 4$.