0
$\begingroup$

enter image description here

In this question $\displaystyle\frac{\log(\sin x)}{\log x}$ is taken as $\displaystyle\frac{\infty}{\infty}$ indeterminate form. But $\log(0)$ is not defined so how can L'Hospital's rule can be used?

$\endgroup$
  • $\begingroup$ Please make some effort when you post your questions, like typing out the question using MathJax $\endgroup$ – Alice Ryhl Oct 2 '14 at 15:58
2
$\begingroup$

Remember that for small $x$ we know $x \approx \sin(x)$. This means that $$\lim_{x \to 0}\frac{\ln(\sin(x))}{\ln(x)} \approx \lim_{x \to 0} \frac{\ln(x)}{\ln(x)}$$ And since $\lim_{x \to 0}\ln(x) =- \infty$, it follows that $\lim_{x \to 0}\frac{\ln(\sin(x))}{\ln(x)}$ is of the form $\frac{-\infty}{-\infty}$.

$\endgroup$
1
$\begingroup$

Because the logarithm function approaches $-\infty$ as its argument approaches $0$ (from the right).

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.