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I have to evaluate the following limit using L'Hospital's rule. $$\lim_{x\to\infty}\frac{\sin(x)}{\sqrt{x^{2}+1}}$$ But when I try to derivate I always get a $\cos(x)$ or $\sin(x)$ function which has no limit when $x\to\infty$.

So, how am I supposed to evaluate it using L'Hospital's rule?

Bernard
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mvfs314
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    You absolutely have to use L'Hopital's? This seems like a prime candidate for the squeeze theorem to me – Stephen Donovan Jun 12 '21 at 22:51
  • @StephenDonovan unfortunately, yes. – mvfs314 Jun 12 '21 at 22:53
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    I'm not even sure how L'Hopital's rule can even be applied, since the function is not an indeterminate form. It's like asking to use L'Hopital to evaluate $$\lim_{x \to \infty} \frac{1}{x}.$$ – heropup Jun 12 '21 at 22:53
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    A bounded function times one that approach to 0 also approaches to zero –  Jun 12 '21 at 22:59
  • This is not the intended way of doing it, but I suppose if you needed to use L’Hopital, you could expand sin(x) into its Maclaurin series. Each term would be some coefficient times a power of x. Use L’hopital’s rule a sufficient number of times for each term, and you get an infinite sum of all zeroes.

    That (while it should work) is mostly a joke. I would not do that as a solution if this is homework.

     – Andrew Sansom Jun 12 '21 at 23:11
  • Alternatively, you could rewrite the quotient by taking the reciprocal of the numerator and put that in the denominator, and similarly the reciprocal of the denominator goes in the numerator. Then you’ll have it written as 0/0, if I did the manipulation correct in my head. The derivatives will be nasty and will require some trig identities, most likely. – Andrew Sansom Jun 12 '21 at 23:16

2 Answers2

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L'Hospital needs to have $\frac{0}{0}$ or $\frac{\infty}{\infty}$, which you have not. On other hand $\left|\frac{\sin(x)}{\sqrt{x^{2}+1}}\right|\leqslant \frac{1}{\sqrt{x^{2}+1}}\to 0$.

Bernard
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zkutch
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  • When the denominator tends to infinity, L'Hôpital's rule in its most general form, does not require the numerator to tend to infinity. It does not even have to have a limit at infinity. Of course, L'Hôpital's rule is not useful in this problem. – Gary Jun 13 '21 at 11:41
  • Good note, @Gary. I think it will be suitable if you point to corresponding link/book. – zkutch Jun 13 '21 at 11:52
  • It is mentioned towards the end of https://en.wikipedia.org/wiki/L%27H%C3%B4pital%27s_rule#General_form and https://en.wikipedia.org/wiki/L%27H%C3%B4pital%27s_rule#Proof_of_L'H%C3%B4pital's_rule – Gary Jun 13 '21 at 11:54
  • zkutch: To add further to Gary's comment about L'Hopital's rule in its general form, you may take a look at here also: https://math.stackexchange.com/questions/4154724/lhospitals-rule-proof-review. You don't need to have $\infty/\infty$ for applicability of L'Hospital's rule. – Koro Jun 18 '21 at 05:24
  • Let me add from my side book Vladimir A. Zorich - Mathematical Analysis I-Springer, 2016, pages 248-249 with general form. – zkutch Dec 09 '23 at 22:38
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Consider $$\lim_{x\to \infty}\frac{\sin x}{\sqrt{x^2+1}}$$ Notice that $$0\leq \frac{\sin x}{\sqrt{x^2+1}} \leq \left|\frac{\sin x}{\sqrt{x^2+1}}\right|\leq \frac{1}{\sqrt{1+x^2}}$$ But when $x\to \infty$ $\frac{1}{\sqrt{x^2+1}}\to 0$

Since $$\lim_{x\to \infty}\frac{1}{\sqrt{1+x^2}}=\lim_{x\to \infty}\frac{\frac{1}{x}}{\sqrt{1+\frac{1}{x^2}}}=0$$ and hence by squeeze theorem $$\lim_{x\to \infty}\frac{\sin x}{\sqrt{x^2+1}}=0$$

Brien Navarro
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