I saw this question few moments back in here. The answer to this question is not 0 and is proceeded using Taylor expansion. I want to know where the error is in the below approach.
$\lim_{x\to0} (\frac{2+cosx}{x^3sinx}−\frac{3}{x^4})$
solving the above as follows:
$=\lim_{x\to0}(\frac{2}{x^3sinx}+\frac{cosx}{x^3sinx}−\frac{3}{x^4})$
$=\lim_{x\to0}(\frac{2}{x^4\frac{sinx}{x}}+\frac{cosx}{x^4\frac{sinx}{x}}−\frac{3}{x^4})$
$=\lim_{x\to0}(\frac{2}{x^4\frac{sinx}{x}}+\frac{1}{x^4\frac{tanx}{x}}−\frac{3}{x^4})$
$=\lim_{x\to0}(\frac{2}{x^4{\lim_{x\to0}\frac{sinx}{x}}}+\frac{1}{x^4\lim_{x\to0}\frac{tanx}{x}}−\frac{3}{x^4})$
$=\lim_{x\to0}(\frac{2}{x^4}+\frac{1}{x^4}−\frac{3}{x^4})$
$=\lim_{x\to0}(\frac{3}{x^4}−\frac{3}{x^4})$
$=\lim_{x\to0}(0)$
$=0$