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.

Printable View

- Nov 18th 2007, 07:04 AMsiriano.3l'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. - Nov 18th 2007, 07:07 AMJhevon
it's a doozie!

by the product rule and chain rule:

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

try seeing what you can simplify before applying L'Hopital's - Nov 18th 2007, 09:32 AMThePerfectHacker
$\displaystyle \lim_{x\to \infty} \frac{\ln (x^4-3)}{\ln x \cos \frac{1}{x}}$ since $\displaystyle \lim_{x\to \infty} \frac{1}{\cos \frac{1}{x}} = 1$. This problem reduces to:

$\displaystyle \lim_{x\to \infty} \frac{\ln (x^4-3)}{\ln x}$.

We can solve this without L'Hopital, note when $\displaystyle x$ is large:

$\displaystyle \ln (x^4) - \ln (3) \leq \ln (x^4 -3) \leq \ln (x^4)$

Thus,

$\displaystyle \frac{\ln x^4 - \ln 3}{\ln x} \leq \frac{\ln (x^4-3)}{\ln x} \leq \frac{\ln x^4}{\ln x}$.

Now the left side and right side limits are 4, because, for example,

$\displaystyle \lim_{x\to \infty} \frac{\ln x^4 - \ln 3}{\ln x} = \lim_{x\to \infty} \frac{\ln x^4}{\ln x} - \underbrace{\frac{\ln 3}{\ln x}}_{\mbox{ this is zero }}$.

But,

$\displaystyle \lim_{x\to \infty} \frac{\ln x^4}{\ln x} = \lim_{x\to \infty} \frac{4\ln x}{\ln x} = 4$.