What assumptions, if any, are required such that the following limit is correct?

$\lim_{x \rightarrow 0^+}{\frac{a \log(x)}{b \log(x)} } = \frac{a}{b}$

  • $\begingroup$ What needed is $b\ne 0$. $\endgroup$ – Ng Chung Tak Oct 21 '16 at 14:23

The only assumption needed is $b \neq 0$.

Since we are approaching 0 from the right, $log(x)$ is defined for all $x>0$, there is no problem there. Thus, the $log(x)$ can be cancelled from bottom and top and we are left with $\dfrac{a}{b}$, which is only defined if $b\neq 0$.


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