Find the limit algebraically :
$$\lim_{x\to 0} \left( \frac{x}{x-\sin x}-\frac{6}{x^2}\right)$$
My Try :
$$\lim_{x\to 0} \frac{x^2}{x^2}\left( \frac{x}{x-\sin x}-\frac{6}{x^2}\right)$$
$$\lim_{x\to 0} \frac{1}{x^2}\left( \frac{x^3}{x-\sin x}-6\right)$$
now:
$$\lim_{x\to 0} \left( \frac{x^3}{x-\sin x}\right)=6$$
So :
$$\lim_{x\to 0} \frac{1}{x^2}\left( \frac{x^3}{x-\sin x}-6\right)=0$$
is it right ?
HINT: use $$\sin x =x-\frac{x^3}{3!}+\frac{x^5}{5!}-...$$so $$\lim_{x\to 0} \left( \frac{x}{x-\sin x}-\frac{6}{x^2}\right)=\\ \lim_{x\to 0} \left( \frac{x}{x-\left(x-\frac{x^3}{3!}+\frac{x^5}{5!}\right)}-\frac{6}{x^2}\right)=\\ \lim_{x\to 0} \left( \frac{x}{\frac{x^3}{3!}-\frac{x^5}{5!}}-\frac{6}{x^2}\right)=\\ \lim_{x\to 0} \left( \frac{1}{\frac{x^2}{3!}-\frac{x^4}{5!}}-\frac{6}{x^2}\right)=\\ \lim_{x\to 0} \frac{6}{x^2}\left( \frac{1}{1-\frac{x^2}{20}}-1\right)=\\$$can you go on ?