Dealing with simple limit of a trigonometric function

Question

I'm dealing with simple function limit. I believe I haven't made any calculating mistakes this far, yet the solution is (wrong) dead end.

$$ \lim_{x \to 0 } \frac{\tan x - \sin x}{\sin^3x} = \Bigg[\frac{1-1}{0}\Bigg] = \Bigg[\frac{0}{0}\Bigg] = \lim_{x \to 0 } \frac{\frac{\tan x}{x}x-\frac{\sin x}{x}x}{\sin^3x} = \lim_{x \to 0 } \frac{x(\frac{\tan x}{x}- \frac{\sin x}{x})}{\sin^3x}$$

I have no idea how to proceed further. What to do with the denominator. Also the numerator in its current state converges to: $$ \big[ 0 (1-1) \big] = \big[0\big]$$ which is problematic since correct solution is $\frac{1}{2}$.

Help's appreciated.

Answer 1

sin x\ne0$$\lim_{x\to0}\dfrac{x\left(\dfrac{\tan x}x-\dfrac{\sin x}x\right)}{\sin^3x}$$ is of the form $$\dfrac00$$ which is Indeterminate .

Try

As $x\to0\sin x\to0\implies \sin x\ne0$

$$\dfrac{\tan x-\sin x}{\sin^3x}=\dfrac{1-\cos x}{\cos x\sin^2x}$$

Again, $$\dfrac{1-\cos x}{\sin^2x}=\dfrac{1-\cos x}{(1-\cos x)(1+\cos x)}=\dfrac1{1+\cos x}$$ for $1-\cos x\ne0\iff\cos x\ne1$ which is true as $x\to0$

Answer 2

From Taylor expansion,

\begin{align}\lim_{x \to 0} \frac{\tan x - \sin x}{\sin^3 x} &= \lim_{x \to 0} \frac{x+\frac13x^3-\left(x-\frac{x^3}{3!} \right)+o(x^3)}{x^3+o(x^3)}\\ &= \lim_{x \to 0} \frac{\left(\frac13+\frac16 \right)x^3}{x^3}\\ &= \frac13 + \frac16\\ &= \frac12\end{align}