Limit calculation

• May 9th 2011, 08:01 AM
moses
Limit calculation
I can't seem to get the hang of calculating these indeterminate forms...how do I figure out

limit ln(cosx) / ln(cos5x) as x goes to pi/2 from the left

The left limits of the numerator and denominator at pi/2 are negative infinity. So I tried looking at the limit of their derivatives: tanx / 5tan(5x) taking out the 1/5 I'm left to calculate the limit of tanx / tan5x. Once again the left limit at pi/2 is infinity.
So looking again at the limit of the derivatives it comes out (secx)^2 / 5(sec5x)^2
which once again I have no idea how to calculate.

I'm very confused.
• May 9th 2011, 08:13 AM
TheEmptySet
Quote:

Originally Posted by moses
I can't seem to get the hang of calculating these indeterminate forms...how do I figure out

limit ln(cosx) / ln(cos5x) as x goes to pi/2 from the left

The left limits of the numerator and denominator at pi/2 are negative infinity. So I tried looking at the limit of their derivatives: tanx / 5tan(5x) taking out the 1/5 I'm left to calculate the limit of tanx / tan5x. Once again the left limit at pi/2 is infinity.
So looking again at the limit of the derivatives it comes out (secx)^2 / 5(sec5x)^2
which once again I have no idea how to calculate.

I'm very confused.

Hint:

$\frac{d}{dx} \ln(cos(x))=\frac{-\sin(x)}{\cos(x)}=-\tan(x)$
• May 9th 2011, 08:42 AM
moses
Okay, so the derivatives of ln(cos(x)) and ln(cos(5x)) are -tan(x) and -5tan(5x). But how do I calculate the limit -tan(x) / -5tan(5x) = (1/5) * (tan(x) / tan(5x))
• May 9th 2011, 08:59 AM
TheEmptySet
Quote:

Originally Posted by moses
Okay, so the derivatives of ln(cos(x)) and ln(cos(5x)) are -tan(x) and -5tan(5x). But how do I calculate the limit -tan(x) / -5tan(5x) = (1/5) * (tan(x) / tan(5x))

$\lim_{x \to 0}\frac{\tan(x)}{5\tan(5x)}=\lim_{x \to 0}\frac{1}{5}\frac{\sin(x)}{\cos(x)}\cdot \frac{\cos(5x)}{\sin(5x)}=\lim_{x \to 0}\frac{1}{5}\frac{\sin(x)}{\sin(5x)}\cdot \frac{\cos(5x)}{\cos(x)}$

$\lim_{x \to 0}\frac{1}{25}\frac{\sin(x)}{x}\cdot \frac{5x}{\sin(5x)}\cdot \frac{\cos(5x)}{\cos(x)}$

Can you finish from here?
• May 10th 2011, 01:06 AM
moses
Yes, I get it now. I wouldn't have thought of splitting up the sinx/5sin5x like that...