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