27
$\begingroup$

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.

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

$\endgroup$
8
  • 3
    $\begingroup$ Try to factor out $x$ $\endgroup$
    – Fakemistake
    Commented Jun 9, 2017 at 6:52
  • 3
    $\begingroup$ It's probably easier to just use squeeze theorem $\endgroup$
    – Brenton
    Commented Jun 9, 2017 at 6:52
  • 23
    $\begingroup$ 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. $\endgroup$
    – Mundron Schmidt
    Commented Jun 9, 2017 at 7:18
  • 3
    $\begingroup$ @AkshayYadav, please consider accepting an answer which solved your problem, there are already many good answers out there, just go and accept one... Cheers :-) $\endgroup$
    – Vidyanshu Mishra
    Commented Jun 19, 2017 at 10:15
  • 1
    $\begingroup$ See math.stackexchange.com/questions/1710786/… $\endgroup$
    – Barry Cipra
    Commented Jun 28, 2017 at 16:16

5 Answers 5

Reset to default
43
$\begingroup$

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

$\endgroup$
2
  • 1
    $\begingroup$ 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? $\endgroup$
    – Akshay Yadav
    Commented Jun 9, 2017 at 7:01
  • 19
    $\begingroup$ @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. $\endgroup$
    – Michael Rozenberg
    Commented Jun 9, 2017 at 9:28
29
$\begingroup$

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.

$\endgroup$
0
17
$\begingroup$

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.

$\endgroup$
0
11
$\begingroup$

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 $1/5$.

$\endgroup$
4
$\begingroup$

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

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .