4
$\begingroup$

So the limit is the following:

$$\lim_{x \to 0}{\frac{x^2-\frac{x^6}{2}-x^2 \cos (x^2)}{\sin (x^{10})}}$$

Expansions for $\sin(x)$ and $\cos(x)$ are given:

$$\sin x = x-\frac{x^3}{3!} + \frac{x^5}{5!}-...+(-1)^{n-1}\frac{x^{2n-1}}{(2n-1)!} + o(x^{2n})$$ $$\cos x = 1-\frac{x^2}{2!}+\frac{x^4}{4!}-...+(-1)^n\frac{x^{2n}}{(2n)!}+o(x^{2n+1})$$

Here is what I tried: $$\lim_{x->0}{\frac{x^2-\frac{x^6}{2}-x^2(1-\frac{x^4}{2}+\frac{x^8}{4!}+o(x^{2*5}))}{x^{10}+o(x^{10*2})}}=\lim_{x->0}{\frac{-\frac{x^{10}}{4!}-o(x^{12})}{x^{10}+o(x^{20})}}$$

This is where I am stuck. I figured that the problem occurs when expending $o()$. What am I missing here?

$\endgroup$
1
  • $\begingroup$ Also better to use \sin than just sin $\endgroup$ – Alex Apr 26 '16 at 21:35
2
$\begingroup$

How about factoring by $x^{10}$?

$\endgroup$
2
  • $\begingroup$ What does $o(x^{20})/x^{10}$ tend to? $\endgroup$ – khajvah Apr 26 '16 at 21:23
  • $\begingroup$ $\frac{o(x^m)}{x^n} $ tends to $0$ in $0$ for $m \geq n \in \mathbb{N}$. Go back to the definition of $o$ to see it $\endgroup$ – Vincent Apr 26 '16 at 21:26
4
$\begingroup$

The numerator is $$x^2-\frac{x^6}2-x^2\Bigl(1-\frac{x^4}2+\frac{x^8}{24}+o(x^8)\Bigr)=-\frac{x^{10}}{24}+o(x^{10}),$$ hence $$\frac{x^2-\cfrac{x^6}{2}-x^2 \cos (x^2)}{\sin (x^{10})}\sim_0\frac{-\dfrac{x^{10}}{24}}{x^{10}}=-\frac1{24}.$$

*Editors Note: Sorry, can't leave comments yet. When you factor in the $-x^2$ you'll end with $-\frac{1}{24}$.

$\endgroup$
2
  • $\begingroup$ Sure, but where is there a term in $\frac1{x^4}$? $\endgroup$ – Bernard Apr 26 '16 at 21:36
  • $\begingroup$ My apologies, I got the term incorrect $\endgroup$ – Alex Apr 26 '16 at 21:38
2
$\begingroup$

Another way is to note that $\frac{\sin x^{10}}{x^{10}} \to_x 1$, so you get your $x^{10}$ in the denominator.

$\endgroup$
3
  • $\begingroup$ The problem still remains in numerator. $o(x^{12})/x^{10}$ doesn't give anything, right? $\endgroup$ – khajvah Apr 26 '16 at 21:27
  • $\begingroup$ $\frac{x^{12}}{x^{10}} = x^2 \to 0$ $\endgroup$ – Alex Apr 26 '16 at 21:28
  • $\begingroup$ Oh god, completely missed that $x->0$... Thanks for help and sorry for wasting your time :) $\endgroup$ – khajvah Apr 26 '16 at 21:29
1
$\begingroup$

$$\lim_{x\to0}\frac{x^2-\frac{x^6}{2}-x^2\cos(x^2)}{\sin(x^{10})} = \lim_{x\to0}\frac{x^2-\frac{x^6}{2}-x^2(1-\frac{x^4}{2}+\frac{x^8}{24}-\frac{x^{12}}{720}+\cdots)}{x^{10}-\frac{x^{30}}{6}+\cdots}$$

$$=\lim_{x\to0}\frac{x^2-\frac{x^6}{2}+(-x^2+\frac{x^6}{2}-\frac{x^{10}}{24}+\frac{x^{12}}{720}-\cdots)}{x^{10}-\frac{x^{30}}{6}+\cdots}$$

Canceling terms in the numerator simplifies the equation to:

$$\lim_{x\to0}\frac{-\frac{x^{10}}{24}+\cdots}{x^{10}-\cdots}$$

The other terms in the series are unnecessary, so

$$\lim_{x\to0}(-\frac{x^{10}}{24}*\frac{1}{x^{10}}) = -\frac{1}{24}$$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.