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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.

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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

3 Answers 3

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.


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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.

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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

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

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