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fine the limit :

$$\lim_{x\to 0}\frac{\sin^2x-x\tan x}{x^4}$$

My Try :

$$\lim_{x\to 0}\frac{\sin^2x-x\tan x}{x^4}\\\lim_{x\to 0}\frac{\frac{\sin x \sin x}{x}-\frac{x\tan x}{x}}{x^4/x}\\\lim_{x\to 0}\frac{\sin x -\tan x}{x^3}=-1/2 $$

is it right ?

user402750
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    No. All the $x$'s have to go to zero at the same time and at the same rate. In your second last line, you can cancel the $x$'s on $\tan x$, but you can't let the $x$ in $\sin x/x$ go to $0$ before the other $x$'s. – B. Goddard Mar 05 '17 at 13:21

2 Answers2

2

HINT:

$$\dfrac{\sin^2x-x\tan x}{x^4}=\dfrac{\sin x}{x\cos x}\cdot\dfrac{\sin x\cos x-x}{x^3}$$

Set $2x=y$ to get $$\lim_{x\to0}\dfrac{\sin x\cos x-x}{x^3}=\lim_{x\to0}\dfrac{\sin2x-2x}{2x^3}=4\cdot\lim_{x\to0}\dfrac{\sin y-y}{y^3}$$

Now see Are all limits solvable without L'Hôpital Rule or Series Expansion

Community
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lab bhattacharjee
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Another hint, write \begin{equation} \frac{\sin^2 x - x\tan x}{x^4} = \frac{\sin^2 x}{x^2}\frac{\sin(2x) - 2x}{x^2\sin(2x)} \end{equation} In the limit $x \rightarrow 0$, $\sin(2x) - 2x = -4x^3/3$. You can confirm using Mathematica that your answer $-2/3$ is indeed correct.

Amey Joshi
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