2
$\begingroup$

Calculate: $$\lim_{x \to 0}\frac{x- \sin{x}}{x^2}$$

I would like to try but i don't find any idea i don't know how to use Hopital rule i tried to return $\cos$ to $\sin$ But it doesn't work a help

$\endgroup$
5
  • $\begingroup$ If there is a limit can you see what it must be? Note that the function is "odd", which means that $f(-x)=-f(x)$. $\endgroup$ – Thomas Andrews Mar 5 '16 at 0:02
  • $\begingroup$ I can't understand $\endgroup$ – user315918 Mar 5 '16 at 0:03
  • 3
    $\begingroup$ Depending on your view, it's perhaps cheating inasmuch as using l'Hopital amounts to using the first terms of a Taylor series, but this follows quickly from the fact that $\sin x = x + O(x^3)$. $\endgroup$ – Travis Willse Mar 5 '16 at 0:05
  • $\begingroup$ it would be more beautiful if instead of $x^2$, you had $x^3$ in denominator. $\endgroup$ – ar2015 Mar 5 '16 at 0:30
  • $\begingroup$ @ar2015, mhd.math's answer can be modified to find the limit with $x^3$ in the denominator (provided you can prove the limit still exists, and assuming you know that $\sin x/x\to1$ as $x\to0$). $\endgroup$ – Barry Cipra Mar 5 '16 at 0:36
2
$\begingroup$

you can prove that the limit exist bu several way

so to find the limit without l'hopital :

$$l=\lim_{x\to 0}\frac{x-\sin x}{x^2}\stackrel{x\to3x}{=}\frac{1}{9}\lim_{x\to 0}\frac{3x-3\sin x+4\sin^3x}{x^2}$$

$$l=\frac{1}{3}\lim_{x\to 0}\frac{x-\sin x}{x^2}+\frac{4}{9}\lim_{x\to 0}\frac{\sin^3x}{x^2}$$

$$l=\frac{1}{3}l+0\Rightarrow l=0$$

$\endgroup$
2
  • $\begingroup$ I like this a lot, but I'd like to see one of the several proofs that the limit exists.... $\endgroup$ – Barry Cipra Mar 5 '16 at 0:34
  • 1
    $\begingroup$ Actually, since $(x-\sin x)/x^2$ is an odd function, the limit, if it exists, can only be $0$, with no fancy trig needed. But I still like the fancy trig! $\endgroup$ – Barry Cipra Mar 5 '16 at 1:08
1
$\begingroup$

The "right" thing to use is the Taylor series, but we will work our way around that. We also avoid L'Hospital's Rule. By the Mean Value Theorem, for any $x\ne 0$ there is a $c_x$ between $0$ and $x$ such that $$\frac{x-\sin x}{x}=1-\cos c_x=2\sin^2(c_x/2).$$ Thus $$\frac{\sin x-x}{x^2}=\frac{2\sin^2(c_x/2)}{x}.$$ Suppose $|x|$ is small non-zero. Since $|c_x/2|\lt |x|/2$ we have $$0\lt \frac{2\sin^2(c_x/2)}{|x|}\lt \frac{x^2/2}{|x|}\lt |x|/2.$$ The right-hand side approaches $0$ as $x\to 0$, so our limit is $0$.

$\endgroup$
0
1
$\begingroup$

L'Hopital is the easy way to do this, but here's another approach, using only $(\sin x)'=\cos x$ and $\sin x\lt x$ (for $x$ positive):

Since ${x-\sin x\over x^2}$ is an odd function, it suffices to consider the one-sided limit with $x\gt0$. We have

$$0\le{x-\sin x\over x^2}={1\over x^2}\int_0^x(1-\cos u)du={1\over x^2}\int_0^x{\sin^2u\over1+\cos u}du\le{1\over x^2}\int_0^x\sin^2udu\\={1\over x^2}\int_0^xu^2{\sin^2u\over u^2}du\le{1\over x^2}\int_0^xu^2du={1\over x^2}\cdot{1\over3}x^3={1\over3}x\to0\text{ as }x\to0^+$$

$\endgroup$
0
$\begingroup$

You can use l'Hopital's rule if you know how to take derivatives. After checking that the fraction is naively $0/0$, differentiate the top of the fraction, and the bottom of the fraction. $$ \frac{1-\cos x}{2x}$$ Still $0/0$? Then do it again: $$ \frac{\sin x}{2} \to 0$$

SO the answer will be zero. Are you sure the problem did not read $$\lim_{x \to 0}\frac{x- \sin{x}}{x^3}$$, which requires three applicatoins and gives an answer of $\frac16$?

$\endgroup$

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