Mathematics - Limits Question with Solution | TestHub

Using Taylor series expansions for $\sin x=x-\frac{x^3}{6}+\frac{x^5}{120}-O(x^7)$ and $\tan x=x+\frac{x^3}{3}+\frac{2x^5}{15}+O(x^7)$, the numerator becomes $a(x-\frac{x^3}{6}+\frac{x^5}{120})-b(x+\frac{x^3}{3}+\frac{2x^5}{15})+c{x}^3$. Grouping terms by powers of $x$: $(a-b)x+(-\frac{a}{6}-\frac{b}{3}+c)x^3+(\frac{a}{120}-\frac{2b}{15})x^5+O(x^7)$. For the limit to be finite and non-zero, the coefficients of $x$ and $x^3$ must be zero. So, $a-b=0⟹a=b$. And $-\frac{a}{6}-\frac{b}{3}+c=0⟹-\frac{a}{6}-\frac{a}{3}+c=0⟹-\frac{a}{2}+c=0⟹c=\frac{a}{2}$. The limit is then the coefficient of $x^5$: $\frac{a}{120}-\frac{2b}{15}=\frac{a}{120}-\frac{2a}{15}=\frac{a-16a}{120}=-\frac{15a}{120}=-\frac{a}{8}$. Given the limit is $\frac{1}{10}$, we have $-\frac{a}{8}=\frac{1}{10}⟹a=-\frac{8}{10}=-\frac{4}{5}$. Thus, $b=-\frac{4}{5}$ and $c=\frac{1}{2}\left(-\frac{4}{5}\right)=-\frac{2}{5}$. The sum $a+b+c=-\frac{4}{5}-\frac{4}{5}-\frac{2}{5}=-\frac{10}{5}=-2$.