Question

Can someone help me please?
How can I tell between infinity and zero for limx->0 for (logcosx)/x ?

Answer 1

Should I generally choose zero or is there a way to tell? The answer is zero when I graph it but I can't tell why it is zero and not infinity.

Answer 2

There are a few things to consider before you find any limits. \(\log x\) is defined for \(x>0\), so \(\log(\cos x)\) is defined for \(\cos x>0\), which is positive for \(-\dfrac{\pi}{2}<x<\dfrac{\pi}{2}\). (This lets us consider the two-sided limit, i.e. there aren't any residual problems with the function's domain.)

So we know that \(\log(\cos x)\) is continuous for \(x\) near 0, which means you can substitute directly:
\[\lim_{x\to0}\frac{\log(\cos x)}{x}=\frac{\displaystyle\lim_{x\to0}\log(\cos x)}{\displaystyle\lim_{x\to0}x}=\frac{\displaystyle\log\left(\lim_{x\to0}\cos x\right)}{\displaystyle\lim_{x\to0}x}=\frac{\log1}{0}=\frac{0}{0}\]
This indeterminate form is appropriate for applying L'Hopital's rule:
\[\lim_{x\to0}\frac{\log(\cos x)}{x}=\lim_{x\to0}\frac{\dfrac{d}{dx}\log(\cos x)}{\dfrac{d}{dx}x}=\lim_{x\to0}\frac{\dfrac{-\sin x}{\cos x}}{1}=\lim_{x\to0}(-\tan x)\]

Answer 3

Thanks... so I need to know L'Hopital's rule and derivatives then. Thanks for answering!