Solve $\lim_\limits{x \to 0} \frac{x^2 \sin x}{\sin x- x \cos x}$

Question

$$ \lim_{x \to 0} \frac{x^2 \sin x}{\sin x - x \cos x} $$ How can one go about solving this limit using real analysis? Would a definition from the calculus proportion of real analysis suffice in order to fully solve this question? What would the proper way of setting out a proof for this problem be? I have tried using the epsilon delta definition which has not worked.

Answer 1

HINT

Use that

• $\sin x=x-x^3/6 +o(x^3)$

• $\cos x=1-x^2/2 +o(x^2)$

then

$$ \lim_{x \to 0} \frac{x^2 \sin x}{\sin x - x \cos x} = \lim_{x \to 0} \frac{x^3+o(x^3)}{-x^3/6 +x^3/2+o(x^3) } $$

Answer 2

You can use l'hospital's rule:

\begin{align} \lim_{x \to 0} \frac{x^2 \sin x}{\sin x - x \cos x} &= \lim_{x \to 0} \frac{2x \sin x+x^2 \cos x}{\cos x - \cos x +x \sin x}\\ &= \lim_{x \to 0} \frac{2x \sin x+x^2 \cos x}{x \sin x}\\ &= \lim_{x \to 0} \frac{2\sin x +2x \cos x +2x \cos x -x^2 \sin x}{ \sin x +x \cos x}\\ &= \lim_{x \to 0} \frac{2\sin x +4x \cos x -x^2 \sin x}{ \sin x +x \cos x}\\ &= \lim_{x \to 0} \frac{2\cos x +4 \cos x -4x \sin x -2x \sin x +x^2 \cos x}{ \cos x + \cos x-x \sin x}\\ &= \frac{2\cos 0 +4 \cos 0 -4\cdot0 \sin0 -2\cdot0 \sin 0 +0^2 \cos 0}{ \cos 0 + \cos 0-0 \sin 0}\\ &= \frac{2 +4 }{ 1 + 1}\\ &= \frac{6}{ 2}=3\\ \end{align}

Answer 3

Alternatively: $$\begin{align}\lim_{x \to 0} \frac{x^2 \sin x}{\sin x - x \cos x}=&\lim_{x \to 0} \frac{x^2 \sin x}{\sin x - x \cos x}\cdot \lim_{x \to 0} \frac{x}{\sin x}=\\ &\lim_{x \to 0} \frac{x^3}{\sin x - x \cos x}\overbrace{=}^{L'H}\\ &\lim_{x \to 0} \frac{3x^2}{x\sin x}=3.\end{align}$$