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I have been recently trying to solve this limit problem. First of all, I used L'Hopital's rule but it doesn't seem to work (because I thought that this limit is of form $\frac{\infty}{\infty}$). Am I doing it correctly? I don't seem to understand where am I wrong.

$$\lim_{x \to \infty} \left(\frac{x+\sin^3x}{5x+6}\right)$$

Michael Rozenberg
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Akshay Yadav
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    Try to factor out $x$ – Fakemistake Jun 09 '17 at 06:52
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    It's probably easier to just use squeeze theorem – Brenton Jun 09 '17 at 06:52
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    To your question why L'Hopital failed: You have to read the statement very carefully! The theorem states, if $\lim_{x\to x_0} f(x)=\lim_{x\to x_0} g(x)\in\{0,\infty\}$ **and** $\lim_{x\to x_0} \frac{f'(x)}{g'(x)}$ exists, **then** there holds $\lim_{x\to x_0} \frac{f(x)}{g(x)}=\lim_{x\to x_0}\frac{f'(x)}{g'(x)}$. It is not said, that this limits are equivalent, but only if the RHS exists, it is the same as the LHS. Your example is a nice counterexample, where the RHS don't exists, while the LHS do. – Mundron Schmidt Jun 09 '17 at 07:18
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    @AkshayYadav, please consider accepting an answer which solved your problem, there are already many good answers out there, just go and accept one... Cheers :-) – Vidyanshu Mishra Jun 19 '17 at 10:15
  • @VidyanshuMishra I don't get what you mean by 'accepting an answer which solved my problem', can you elaborate? – Akshay Yadav Jun 20 '17 at 14:00
  • There is a check mark on the left hand of each answer, if you click on that it means you have accepted that answer. You accept an answer which you like the most.click – Vidyanshu Mishra Jun 20 '17 at 16:59
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    See https://math.stackexchange.com/questions/1710786/why-does-lhopitals-rule-fail-in-this-case – Barry Cipra Jun 28 '17 at 16:16
  • Does this answer your question? [Why does L'Hopital's rule fail in calculating $\lim_{x \to \infty} \frac{x}{x+\sin(x)}$?](https://math.stackexchange.com/questions/1710786/why-does-lhopitals-rule-fail-in-calculating-lim-x-to-infty-fracxx-s) – Ashish Ahuja May 14 '21 at 10:45

5 Answers5

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$$\lim_{x\rightarrow\infty}\frac{x+\sin^3x}{5x+6}=\lim_{x\rightarrow\infty}\frac{1+\frac{\sin^3x}{x}}{5+\frac{6}{x}}=\frac{1}{5}$$

Michael Rozenberg
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    Sir, how did you concluded that $\lim_{x \to \infty} (\sin^3(x)/x)=0$. Moreover can you tell me why did L'Hospital rule failed? – Akshay Yadav Jun 09 '17 at 07:01
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    @Akshay Yadav Because $-1\leq\sin^3x\leq1$, which gives $-\frac{1}{|x|}\leq\frac{\sin^3x}{x}\leq\frac{1}{|x|}$. The L'Hospital rule failed because $\lim\limits_{x\rightarrow\infty}(\sin^3x)'$ does not exist. – Michael Rozenberg Jun 09 '17 at 09:28
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L'Hospital's Rule isn't working since the derivative of numerator function isn't determinable when $x \to \infty$, due to oscillatory behaviour of $\sin$ and $\cos$ function. Therefore you have to approach traditionally.

Jaideep Khare
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We can squeeze $(x+\sin^3x)/(5x+6)$ between two applications of l'Hopital: When $5x+6>0$ we have $$\frac {x-1}{5x+6}\leq \frac {x+\sin^3x}{5x+6}\leq \frac {x+1}{5x+6}.$$ Applying l'Hopital to the far left and far right of the above line, we see they both have limits of $1/5$ as $x\to \infty.$ So the expression in the middle must also go to $1/5$.

Of course we could also re-write the far L & far R as $\frac {1}{5}(1- \frac {11}{5x+6})$ and $\frac {1}{5}(1-\frac {1}{5x+6})$ and not need l'Hopital.

DanielWainfleet
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One of the conditions of applying L'Hospital's Rule is that $f'(x)/g'(x)$ must exist. $$\lim\limits_{x \to \infty} \frac{f'(x)}{g'(x)}$$

After one application of L'Hospitals, you arrived at a finite numerator over a finite denominator. But while the numerator was finite, it was non-convergent, and so that limit did not exist. Just like the much simpler $\sin(x)$ does not converge to a single value when x approaches infinity -- it oscillates between +/-1. $$\lim\limits_{x \to \infty} {\sin(x)}$$

So all of L'Hospital's pre-conditions must exist for you to use it. As others have mentioned, this limit could more easily be solved by using the squeezing theorem. The numerator's value gets squeezed between $x+1$ and $x-1$. Both of those limits go to $1/5$.

Jaideep Khare
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John
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For some limits L'Hospital rule doesn't seem to work ,like what you write $$\lim_{x \to \infty} \left(\frac{x+\sin^3(x)}{5x+6}\right)\\\lim_{x \to \infty} \left(\frac{3x+\cos(x)}{x+6\sin(2x)}\right)\\\lim_{x \to \frac{\pi}{2}} \left(\frac{\tan x}{\tan (3x)}\right)\\\lim_{x \to 0} \left(\frac{\cot (2x)}{x^2+\cot x}\right)\\\vdots$$ but (all i see ) have alternative solution ,or change something you may use L'Hospital rule . $$\lim_{x \to \infty} \left(\frac{x+\sin^3(x)}{5x+6}\right)=\lim_{x \to \infty} \left(\frac{x(1+\frac{\sin^3(x)}{x})}{x(5+\frac6x)}\right)=\\\lim_{x \to \infty} \left(\frac{(1+\frac{\sin^3(x)}{x})}{(5+\frac6x)}\right)=\frac15$$ for $\\\lim_{x \to \frac{\pi}{2}} \left(\frac{\tan x}{\tan (3x)}\right)$ we have $$\lim_{x \to \frac{\pi}{2}} \left(\frac{\tan x}{\tan (3x)}\right)=\\\lim_{x \to \frac{\pi}{2}} \left(\frac{\frac{1}{\cot x}}{\frac{1}{\cot 3x}}\right)=\\\lim_{x \to \frac{\pi}{2}} \left(\frac{\cot 3x}{\cot x}\right)$$ now you can use L'Hospital rule

I think when you have a trig function $\to \infty $ L'Hospital rule is not suitable

Khosrotash
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