Can you verify the solution to the following (very simple) problem? $$\lim_{x\to 0^+} \frac{\ln(\sin(x/2))}{\ln(\sin x)}.$$
$$\text{limit} = \lim_{x\to 0^+} \frac{\ln(\frac{\sin(x/2)}{x/2})+\ln(x/2)}{\ln(\sin x/x)+\ln(x)} = \lim_{x\to 0^+} \frac{\frac{\ln(\frac{\sin(x/2)}{x/2})}{\ln(x)}+\frac{\ln(x/2)}{\ln(x)}} {\frac{\ln(\sin x/x)}{\ln(x)}+1} \\= \lim_{x\to 0^+} \frac{\frac{\ln 1}{-\infty}+\frac{\ln(x/2)}{\ln (x)}}{\frac{\ln 1}{-\infty}+1} = \lim_{x\to 0^+} \frac{\ln(x/2)}{\ln(x)} = \lim_{x\to 0^+} \frac{\ln(x)-\ln 2}{\ln(x)} =1. $$
What did I do wrong? The given answer is $2$.
It would be better if you solved your exercise in the following way:
$$\text{limit}=\lim_{x\to 0^+}\frac{\ln\left(\frac{\sin(x/2)}{x/2}\right)+\ln(x/2)}{\ln\left(\frac{\sin x}x\right)+\ln(x)}=\lim_{x\to 0^+} \frac{\frac{\ln\left(\frac{\sin(x/2)}{x/2}\right)}{\ln(x)}+\frac{\ln(x/2)}{\ln(x)}}{\frac{\ln\left(\frac{\sin x}x\right)}{\ln(x)}+1}$$
$$=\lim_{x\to 0^+}\frac{\frac{\ln\left(\frac{\sin(x/2)}{x/2}\right)}{\ln(x)}+\frac{\ln(x)-\ln2}{\ln(x)}}{\frac{\ln\left(\frac{\sin x}x\right)}{\ln(x)}+1}=\lim_{x\to 0^+}\frac{\frac{\ln\left(\frac{\sin(x/2)}{x/2}\right)}{\ln(x)}+1-\frac{\ln2}{\ln(x)}}{\frac{\ln\left(\frac{\sin x}x\right)}{\ln(x)}+1}$$
$$=\frac{\frac{\ln1}{-\infty}+1-\frac{\ln2}{-\infty}}{\frac{\ln1}{-\infty}+1}= \frac11=1. $$
In general, it is not correct to evaluate the limit of only a part of the expression. You have to evaluate the limit of the whole expression.