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What's the limit of: $\lim_{x \rightarrow 0}\frac{sin(x)+cos(x)}{x}$ ?

$\lim_{x \rightarrow 0} \left (sin(x) + cos(x) \right) = sin(0)+cos(0) = 1 $

$\lim_{x \rightarrow 0} x = 0$

$\Rightarrow \frac{1}{0} \Rightarrow$ L'Hôpital's rule is needed

$f'(x) = cos(x) - sin(x)$

$g'(x) = 1$

$\Rightarrow $

$\lim_{x\rightarrow 0} \left (cos(x) - sin(x) \right ) = cos(0) -sin(0) = 1-0 = 1$

So 1 will be the limit? No way because as it looks like, the denumerator will be too small and thus the complete function will go towards $\infty$ for $x \rightarrow 0$.

This is very confusing for me :S

Maybe the mistake is using L'Hôpital?

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    $\begingroup$ You can't use Lhopital here. $\endgroup$ – user223391 Jul 23 '16 at 16:46
  • $\begingroup$ Thought so... But why not? $\endgroup$ – cnmesr Jul 23 '16 at 16:47
  • $\begingroup$ Because $1/0$ is not an indeterminate form. Here, this limit does not exist, which can be verified by graphing (please do as part of the learning process) $\endgroup$ – imranfat Jul 23 '16 at 16:48
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    $\begingroup$ $1/0$ indicates not a need for l'Hôpital, but a diverging result. l'Hôpital is for $0/0$ and $\infty/\infty$. $\endgroup$ – celtschk Jul 23 '16 at 16:49
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    $\begingroup$ You are missing all the intuition here. What is $1/(.00000000001)?$ $\endgroup$ – zhw. Jul 23 '16 at 17:38
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l'Hopital's rule only works for the limits "$0/0$" and "$\infty/\infty$".

To compute this limit try this: $$\frac{\sin x+\cos x}x=\frac{\sin x}x+\frac{\cos x}x$$

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  • $\begingroup$ The numerator has limit $1$, the denominator has limit $0$; no need to split: the limit is $\infty$ for $x\to0^+$ and $-\infty$ for $x\to0^-$. $\endgroup$ – egreg Jul 23 '16 at 16:51
  • $\begingroup$ Doesn't seem to help: $\lim_{x\rightarrow 0}\frac{sin(x)}{x} = \frac{0}{0}$ and $\lim_{x\rightarrow 0}\frac{cos(x)}{x} = \frac{1}{0}$ can use L'Hôpital for first but not for second... $\endgroup$ – cnmesr Jul 23 '16 at 16:53
  • $\begingroup$ @cnmesr L'Hopital is not meant to be used on this problem, the limit does not exist either. $\endgroup$ – Will Fisher Jul 23 '16 at 16:55
  • $\begingroup$ But how would you write that down then? As I did in the comment now? The first will go to 1 and the other to $\infty$, so in total it will all go to $\infty$..? $\endgroup$ – cnmesr Jul 23 '16 at 16:56
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    $\begingroup$ No, I would say DNE, which means does not exist. $\endgroup$ – imranfat Jul 23 '16 at 17:00
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Because the question is not in indeterminate form i.e. it's not $\frac{0}{0}$.

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  • $\begingroup$ There is a little more that that though...(didn't downvote) $\endgroup$ – imranfat Jul 23 '16 at 16:50
  • $\begingroup$ I give up vote so it's normal again. Not fair to negate someone for trying to help imo. $\endgroup$ – cnmesr Jul 23 '16 at 17:01

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