How to find the limit $\lim_{x\to0}\frac{x\tan x-x\sin x}{x\sin^2x/\cos x}$

Question

Here is the limit I'm struggling with: $$\lim_{x\to0}\cfrac{x\tan x-x\sin x}{x\sin^2x/\cos x}.$$ Worked so hard to find it, but couldn't.

Answer 1

Hint:
First notice that: $$\lim_{x\to0}\dfrac{x\tan x-x\sin x}{x\sin^2x/\cos x}=\lim_{x\to0}\dfrac{x(\tan x-\sin x)}{x\sin^2x/\cos x}=\require{cancel}\lim_{x\to0}\dfrac{\cancel x(\tan x-\sin x)}{\cancel x\sin^2x/\cos x}.$$ Then you can simplify it further as follows: $$\lim_{x\to0}\dfrac{\tan x-\sin x}{\sin^2x/\cos x}=\lim_{x\to0}\dfrac{\tan x-\sin x}{\sin x(\sin x/\cos x)}=\lim_{x\to0}\dfrac{\tan x-\tan x\cos x}{\sin x\tan x}.$$ Factor $\tan x$ and now try to identify expressions involving those two known limits: $$\lim_{x\to0}\dfrac{1-\cos x}x=0\quad\color{grey}{\rm and}\quad\lim_{x\to0}\dfrac{x}{\sin x}=1.$$

Answer 2

rewrite $\tan x = \frac{\sin x}{\cos x}$ and use the limit $\frac{\sin x }{x} \to_{x} 1$ and also $\lim_{x \to 0} \cos x = 1$

Answer 3

Simplify by $x\sin x$ and multiply numerator and denominator by $\cos x$ the limit becomes

$$\lim_{x\to0}\frac{1-\cos x}{\sin x}=\lim_{x\to0}\frac{x^2}{2 x}=0$$

Answer 4

Multiply top and bottom by $(\cos x) / x$:

$$\lim_{x\to0}\frac{\sin x-\sin x \cos x}{\sin^2x} = \lim_{x\to0}\frac{(\sin x)(1- \cos x)}{\sin^2x} = \lim_{x\to0}\frac{1- \cos x}{\sin x}$$

Now divide top and bottom by $x$:

$$=\lim_{x\to0}\frac{\frac{1- \cos x}{x}}{\frac{\sin x}{x}}$$

The limit of the top & bottom should be familiar to you:

$$\lim_{x\to0}\frac{\frac{1- \cos x}{x}}{\frac{\sin x}{x}} = \frac{0}{1} = 0$$

Answer 5

I hope it's not too late...

$\dfrac{x\tan x - x\sin x}{\dfrac{x\sin ^2x}{\cos x}} \sim_0 \cos x⋅ \dfrac{x\left(x + \dfrac{x^3}{3}\right) - x\left(x - \dfrac{x^3}{6}\right)}{x\left(x^2 - \dfrac{x^4}{3}\right)} \sim_0 1⋅ \dfrac{3x}{6 - 2x^4} → 0$ as $x → 0$. The answer follows.