$\lim_{x\to 0}\frac{2+\cos x}{x^3\sin x}-\frac{3}{x^4}=\frac{1}{60}$ without using L Hospital rule or series expansion

Question

Prove that $\lim_{x\to 0}\frac{2+\cos x}{x^3\sin x}-\frac{3}{x^4}=\frac{1}{60}$ without using L Hospital rule or series expansion.

I tried $\lim_{x\to 0}\frac{2+\cos x}{x^3\sin x}-\frac{3}{x^4}=\lim_{x\to 0}\frac{2+\cos x}{x^4\lim_{x\to 0}\frac{\sin x}{x}}-\frac{3}{x^4}$
$=\lim_{x\to 0}\frac{2+\cos x}{x^4}-\frac{3}{x^4}=\lim_{x\to 0}\frac{\cos x-1}{x^4}=\lim_{x\to 0}\frac{-2\sin^2 \frac{x}{2}}{x^4}=\lim_{x\to 0}\frac{-2\sin^2 \frac{x}{2}}{4x^2(\frac{x}{2})^2}$
$=\lim_{x\to 0}\frac{-1}{2x^2}$
Now it has turned into limit does not exist.I dont know where have i made mistake,because as per my knowledge my every step is correct.If i have made mistake please correct me
Please help me.Thanks.

Related : http://math.stackexchange.com/questions/387333/are-all-limits-solvable-without-lh%C3%B4pital-rule-or-series-expansion — lab bhattacharjee, Dec 19 '15 at 03:41
I was reading the post suggested by you,but wondering whether my question can be solved by the method suggested in this.@labbhattacharjee — diya, Dec 19 '15 at 03:43
Your mistake is probably in the first step when you took the limit of $\sin{x}/x$ inside the big limit. This isn't allowed since the limit of $\frac{2+\cos x}{x^4}$ doesn't exist. — Joey Zou, Dec 19 '15 at 03:47
Yes i got the answer by method suggested by @labbhattacharjee.Thank you. — diya, Dec 19 '15 at 03:49
@diya, you can answer your own question, always appreciated by the community! — Jesse P Francis, Dec 19 '15 at 03:51

score 2 · Answer 1 · edited Dec 19 '15 at 05:49

2

As per the method suggested by Lab bhattacharjee, I got the answer to my problem.Thanks to him.

Let $L=\lim_{x\to 0}\frac{2+\cos x}{x^3\sin x}-\frac{3}{x^4}......(1)$

So $L=\lim_{x\to 0}\frac{2+\cos 2x}{8x^3\sin 2x}-\frac{3}{16x^4}$

$L=\lim_{x\to 0}\frac{1+2\cos^2x}{16x^3\sin x\cos x}-\frac{3}{16x^4}$

$16L=\lim_{x\to 0}\frac{1+2\cos^2x}{x^3\sin x\cos x}-\frac{3}{x^4}......(2)$

Subtracting $(1)$ from $(2)$,we get
$15 L=\lim_{x\to 0}\frac{1-2\cos x+\cos^2 x}{x^3\sin x}$
$15 L=\lim_{x\to 0}\frac{(1-\cos x)^2}{x^4\lim_{x\to 0}\frac{\sin x}{x}}$
$15L=\lim_{x\to 0}\frac{4\sin^4\frac{x}{2}}{x^4}$
$15L=\lim_{x\to 0}\frac{4\sin^4\frac{x}{2}}{16(\frac{x}{2})^4}$
$L=\frac{1}{60}\lim_{x\to 0}(\frac{\sin\frac{x}{2}}{\frac{x}{2}})^4$
$L=\frac{1}{60}$

edited Dec 19 '15 at 05:49

Harish Chandra Rajpoot

37,450

answered Dec 19 '15 at 05:35

diya

3,589

You've shown if the limit exists, it equals $1/60.$ but you haven't shown the limit exists. – zhw. Dec 19 '15 at 06:49
How can i show that limit exists.I dont know.Can you show me please?@zhw. – diya Dec 19 '15 at 07:15

$\lim_{x\to 0}\frac{2+\cos x}{x^3\sin x}-\frac{3}{x^4}=\frac{1}{60}$ without using L Hospital rule or series expansion

1 Answers1