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 some one 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 '13 at 18:39
  • $\begingroup$ @amWhy I fixed it. It was just a .png. For OP: Try dividing everything by $x$ and recalling that $\lim_{x\rightarrow 0}\frac{\sin x}{x} = 1$. $\endgroup$ – Cameron Williams Nov 26 '13 at 18:39
  • $\begingroup$ yeah thanks this site is a bit of a pain sorry about that $\endgroup$ – wolfcall Nov 26 '13 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 '14 at 14:28

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 '13 at 18:43
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    $\begingroup$ I've updated, wolfcall, now that the original question has been corrected! ;-) $\endgroup$ – Namaste Nov 26 '13 at 18:52
  • $\begingroup$ @amWhy: Needs another TU +1 $\endgroup$ – Amzoti Nov 27 '13 at 3:52

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

  • $\begingroup$ Thank you thank you thank you $\endgroup$ – wolfcall Nov 26 '13 at 18:47
  • $\begingroup$ No problem! You should accept one of the answers (either mine or amWhy's). $\endgroup$ – Dustan Levenstein Nov 26 '13 at 18:48

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