Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I want to prove the following limit $\lim_{x\rightarrow 0} \frac{\sin x}{x^{2}}=\infty$ using the definition of limit. I have used the inequality $\left|\sin x \right|\geq \left|x-\frac{x^{3}}{6}\right|$ in a neighborhood of 0. It is not wrong, but $x-\frac{x^{3}}{6}$ is the third-order Taylor polynomial for $\sin x$. The inequality also can be easily demonstrated without the use of Taylor polynomials. However, I would like to find a different solution, which does not use the Taylor polynomials.

share|cite|improve this question
I assume you're looking for a proof that the limit from the right is $\infty$. From the left the limit is $- \infty$. – Patrick Mar 7 '12 at 15:15
@Mark One has that $\sin x \sim x$, so that your limit will ultimately behave like $\lim\limits_{x \to 0} \frac{1}{x} $ – Pedro Tamaroff Mar 7 '12 at 16:18
@Patrick: yes, it is. – Mark Mar 7 '12 at 19:48
up vote 3 down vote accepted

You can use the standard geometric argument that for $0<t<{\pi\over2}$ $$ \cos t \le {\sin t\over t}\le1. $$ From this it follows that for $t$ sufficiently small: $$ \Bigl|{\sin t\over t^2} \Bigr|\ge\Bigr|{\cos t\over t}\Bigr|\ge{1\over 2|t|}. $$

[Edit] But, note Patrick's comment.

enter image description here

share|cite|improve this answer
and $\left|\cos t \right|\geq 1/2 $ because you consider the neighborhood of 0 in the interval $(-\pi/3, \pi/3)$... thank you very much. – Mark Mar 7 '12 at 19:57

You can try proving that $|\sin x| \ge \left|\frac{x}{2}\right|$ in a neigbourhood of zero.

One way to prove that is to consider the $\epsilon-\delta$ definition of this limit:

$$\lim_{x \to 0} \frac{\sin x}{x} = 1$$

and choose an appropriate $\epsilon$.

Another way is to just consider the function $f(x) = 2\sin x - x$ and investigate its properties near $0$.

btw, you don't need the Taylor series to prove $|\sin x| \ge |x - x^3/6|$, the second idea above can be used here too.

share|cite|improve this answer
Yes, I don't need the Taylor series to prove the inequality, but I wanted an alternative method, purely geometric. Thank you very much. – Mark Mar 7 '12 at 19:52

You can prove that some of the limit laws hold for a sequence $a_{n}\to\infty$ and $b_{n}\to L$. Then it is immediate from

$$\lim_{x\to0^{+}}\frac{\sin x}{x}\frac{1}{x}=``1 \cdot\infty "= \infty$$

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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