# Thread: Having trouble evaluating this limit

1. ## Having trouble evaluating this limit

y = lim [X->0] (ln x)/ ln ( cos x)

can't use L'Hopital's rule since the indeterminant form is infinity/0. you can use it only when it's 0/0 or infinity/infinity.

the final answer is positive infinity, but i don't know how to get there. been at it for days

Thanks

2. limit as x goes to 0 of - Wolfram|Alpha[%28Log[x]%29%2F%28Log[Cos[x]]%29]

Click show steps.

3. Originally Posted by eddie2042
y = lim [X->0] (ln x)/ ln ( cos x)

can't use L'Hopital's rule since the indeterminant form is infinity/0. you can use it only when it's 0/0 or infinity/infinity.

the final answer is positive infinity, but i don't know how to get there. been at it for days

Thanks

The limit must be $\lim_{x\to 0^+}\frac{\ln x}{\ln\cos x}$ , since the logarithm is defined only on the positive reals, but then the numerator goes to $-\infty$ whereas the denominator

goes to $\ln 1=0$ from the left since $|\cos x| \leq 1$ , so it also is negative all the time.

Tonio

4. any time the denominator goes to zero or the numerator goes to infinity the expression goes to infinity

Similarly any time the numerator goes to zero or the denominator goes to infinity the expression goes to zero

The only exception is when both numerator and denominator approach infinity or zero