l'Hôpital's Rule

siriano.3 · November 18th 2007, 07:04 AM

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.

**Jhevon** · November 18th 2007, 07:07 AM

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

**ThePerfectHacker** · November 18th 2007, 09:32 AM

$\lim_{x\to \infty} \frac{\ln (x^4-3)}{\ln x \cos \frac{1}{x}}$ 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,
$\lim_{x\to \infty} \frac{\ln x^4}{\ln x} = \lim_{x\to \infty} \frac{4\ln x}{\ln x} = 4$ .

Thread: l'Hôpital's Rule

LinkBack

Thread Tools

Display

l'Hôpital's Rule

Similar Math Help Forum Discussions

L' Hopital's Rule

L'Hopital's Rule

l’Hopital’s Rule Help..

L'hopital Rule qn

L'Hopital Rule