This is an assignment and I am stuck:

Find the limit, whether finite or infinite, or indicate that the limit does not exist. Use l'Hôpital's Rule if appropriate.

$$\lim_{x\rightarrow 0} \frac{4x+4\sin x}{10x+10\cos x}.$$

When I try l'Hôpital's rule I get $\frac 8{10}$. Can someone tell me why this isn't right?

  • $\begingroup$ sorry I will type it when it preview it showed the image so i thought it would work $\endgroup$
    – wolfcall
    Nov 26, 2013 at 18:39
  • $\begingroup$ @amWhy I fixed it. It was just a .png. For OP: Try dividing everything by $x$ and recalling that $\lim_{x\rightarrow 0}\frac{\sin x}{x} = 1$. $\endgroup$ Nov 26, 2013 at 18:39
  • $\begingroup$ yeah thanks this site is a bit of a pain sorry about that $\endgroup$
    – wolfcall
    Nov 26, 2013 at 18:40
  • $\begingroup$ If the limit of the numerator f (x) is 0 (inf) and the limit of the denominator g (x) is also 0 (inf) and the limit of f' (x) / g' (x) exists, then the limit of f (x) / g (x) exists and is the same. Which of the three conditions fails? $\endgroup$
    – gnasher729
    Apr 28, 2014 at 14:28
  • $\begingroup$ Check again the prerequisities to l'Hôpital's rule. Do expressions above and below the fraction line satisfy them? And – do you really need the rule at all? How about just calculating the subexpressions? $\endgroup$
    – CiaPan
    Jan 5, 2021 at 14:51

3 Answers 3


Original posted question:

$$\lim_{x\rightarrow 0} \frac{4x+4\sin x}{10x+10\sin x} = \lim_{x\to 0} \frac{(4x + 4\sin x)'}{(10x + 10 \sin x)'} = \lim_{x\to 0} \frac{4+4\cos x}{10 + 10\cos x} = \frac {4 + 4}{10 + 10} = \frac {8}{20} = \frac 25$$

Since you meant to post $$\lim_{x\rightarrow 0} \frac{4x+4\sin x}{10x+10\cos x}$$ note that in this case, l'Hôpital is not applicable (the limit does not at first evaluate to an indeterminate limit). Nor would we want to use it! It is easily solved by evaluating immediately:

$$\lim_{x\rightarrow 0} \frac{4x+4\sin x}{10x+10\cos x} = \frac{ 0 + 0}{0 + 10(1)} = \frac {0}{10} = 0$$

IMPORTANT TO REMEMBER: We apply l'Hôpital's rule if and only if a limit evaluates to an indeterminate form. That bold-face link will take you to Wikipedia's concise list of "what counts" as an indeterminate form, and why.

  • 1
    $\begingroup$ the bottom was cos not sin $\endgroup$
    – wolfcall
    Nov 26, 2013 at 18:43
  • 1
    $\begingroup$ I've updated, wolfcall, now that the original question has been corrected! ;-) $\endgroup$
    – amWhy
    Nov 26, 2013 at 18:52

Just plug in! The denominator is $10$, not zero. ;)

  • $\begingroup$ Thank you thank you thank you $\endgroup$
    – wolfcall
    Nov 26, 2013 at 18:47
  • $\begingroup$ No problem! You should accept one of the answers (either mine or amWhy's). $\endgroup$ Nov 26, 2013 at 18:48

It may be a non-rigorous approach but here you go :

When $x\to 0$ we have these two approximations :

$$\sin x \overset{x\to 0}\sim x \ \ \ \ \ \text{and}\ \ \ \ \ \cos(x) \overset{x\to 0}\sim 1-\frac{x^2}{2}$$ Let :

$$\mathcal{L} =\lim_{x\rightarrow 0} \frac{4x+4\sin x}{10x+10\cos x}$$ Using those approximations we get : \begin{align} \mathcal{L}&=\lim_{x\to 0} \frac{4x+4x}{10x-10-5x^2}\\ &=\lim_{x\to 0} \frac{8x}{-5x^2+10x-10}\\ &= \frac{0}{10}\\ &=0 \end{align} Your limit doesn't need anything it's already equals $0$ without direct substitutions.


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