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$\displaystyle \lim_{x \to \infty} \frac{\cos{x}}{x^2} =\lim_{x \to \infty} \frac{\frac{d}{dx} \cos{x}}{\frac{d}{dx}x^2} = -\frac{1}{2}\lim_{x \to \infty}\frac{\sin{x}}{x}$. But $-\frac{1}{x} \le \frac{\sin{x}}{x} \le \frac{1}{x}$ so $\displaystyle -\lim_{x \to \infty}\frac{1}{x} \le \lim_{x \to \infty}\frac{\sin{x}}{x} \le \lim_{x \to \infty}\frac{1}{x} \iff 0 \le \lim_{x \to \infty}\frac{\sin{x}}{x} \le 0 \iff \lim_{x \to \infty}\frac{\sin{x}}{x} = 0.$

Therefore $\displaystyle \lim_{x \to \infty} \frac{\cos{x}}{x^2} = 0.$

Is the above correct?

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    $\begingroup$ You can use the bound $\frac{-1}{x^{2}} \leqslant \frac{\cos x}{x^{2}} \leqslant \frac{1}{x^{2}}$ to start. $\endgroup$ Jan 1 '16 at 0:47
  • $\begingroup$ Your reasoning is correct but more complicated than it needs to be. $|\cos x|\le 1$, $\lim 1/x^2=0$ suffices. $\endgroup$
    – ForgotALot
    Jan 1 '16 at 0:47
  • $\begingroup$ You can only use L'Hopital if it is $0/0$ or $\infty/\infty$. This is neither. $\endgroup$
    – Empy2
    Jan 1 '16 at 0:48
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    $\begingroup$ @Michael I have checked Baby Rudin 5.13. Per that section, L'Hopital's rule applies to anything$/\pm\infty$. Thus, the reasoning above still seems correct to me, just overly complicated. $\endgroup$
    – ForgotALot
    Jan 1 '16 at 15:23
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You must not apply L'Hospital's rule in this case! L'Hospital is only applicable for limits with $\infty \over \infty$ or $0 \over 0$ expressions.

Nonetheless, you can easily get to the limit knowing that $\mid\cos x \mid \leq 1$, thus $$\frac{\mid\cos x\mid}{x^2} \leq \frac{1}{x^2} \xrightarrow{\: n \to \infty \: } 0$$

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  • $\begingroup$ Good catch and Well answered. $\endgroup$
    – user174622
    Jan 1 '16 at 0:54
  • $\begingroup$ Good catch! (+1) $\endgroup$ Jan 1 '16 at 0:54
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You arrived at the correct answer, but your first step is incorrect.

Here is another method:

Note that $\cos(x)$ is bounded and $$\lim_{x\rightarrow \infty}\frac{1}{x^2}=0$$ And you're done.

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  • $\begingroup$ how is l'hospital correct? Is it $0 \over 0$ or $\infty \over \infty$? $\endgroup$
    – adjan
    Jan 1 '16 at 0:51
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    $\begingroup$ @addy2012 You are absolutely correct. Fixed. $\endgroup$
    – user174622
    Jan 1 '16 at 0:51
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$$ \left|\frac{\cos x}{x^2}\right|\le \frac{1}{x^2}\to 0. $$

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You don’t need to use the derivatives: For every $x > 0$ we have $$ -\frac{1}{x^2} \leq \frac{\cos(x)}{x^2} \leq \frac{1}{x^2}. $$ Because $\lim_{x \to \infty} \frac{1}{x^2} = 0$ it follows that $\lim_{x \to \infty} \frac{\cos(x)}{x^2} = 0$.

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L'hospital rule is not correct in this case as noticed by @addy2012.

You may observe that you could conclude directly, as $x \to \infty$, by the squeeze theorem: $$ \left|\frac{\cos x}{x^2}\right|\leq \frac1{x^2}. $$

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  • $\begingroup$ usage of l'hospital is not correct... $\endgroup$
    – adjan
    Jan 1 '16 at 0:50
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    $\begingroup$ @addy2012 Oops, you are right! I will edit my answer, thanks! $\endgroup$ Jan 1 '16 at 0:52
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Contrary to many other answers (including the accepted answer by adjan) your method is correct. L'Hospital's Rule is applicable for indeterminate forms like $0/0$ and "$\text{anything}/\pm\infty$".

However the use of L'Hospital's Rule is totally unnecessary here and the limit is evaluated more easily by applying "Squeeze theorem" on the inequality $$-\frac{1}{x^{2}}\leq \frac{\cos x}{x^{2}} \leq \frac{1}{x^{2}}$$

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