Find lim as x to 0+ (ln (sin mx))/(ln sin x), where m>0

Find $\displaystyle\lim_{x\to0^{+}}\frac{\ln\bigl(\sin mx\bigr)}{\ln\sin x}$, where $m>0$.

Step 1: The form

As $x\to0^{+}$ both $\sin mx$ and $\sin x$ tend to $0^{+}$, so both logarithms tend to $-\infty$: the form is $\dfrac{\infty}{\infty}$.

Step 2: Separate the divergent part

Write each sine as a ratio times the variable, so that the logarithm splits: \[\ln\bigl(\sin mx\bigr) = \ln\frac{\sin mx}{mx}+\ln m+\ln x, \qquad \ln\bigl(\sin x\bigr) = \ln\frac{\sin x}{x}+\ln x.\]

Only the term $\ln x$ is unbounded; everything else tends to a finite value, since $\dfrac{\sin t}{t}\to1$ gives $\ln\dfrac{\sin t}{t}\to0$.

Step 3: Divide by $\ln x$

\[ \frac{\ln\bigl(\sin mx\bigr)}{\ln\sin x} = \frac{\dfrac{\ln\frac{\sin mx}{mx}}{\ln x}+\dfrac{\ln m}{\ln x}+1}{\dfrac{\ln\frac{\sin x}{x}}{\ln x}+1}. \]

Each fraction has a bounded numerator over $\ln x\to-\infty$, so each tends to $0$.

Step 4: Verification

\[\text{By l'H\^opital: } \frac{m\cot mx}{\cot x} = \frac{m\tan x}{\tan mx} \to \frac{m\cdot x}{mx} = 1 \;\;\checkmark\]

\[m = 3,\ x = 10^{-3}:\ \frac{\ln0.0029999}{\ln0.001} = \frac{-5.8091}{-6.9078} = 0.84095\]

\[m = 3,\ x = 10^{-10}:\ \frac{-22.8}{-23.0} = 0.9523 \ \text{— the approach is slow, like } 1+\frac{\ln m}{\ln x} \;\;\checkmark\]