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This is the limit: $$ \lim_{x\to 0}\frac{\sin(x)}{\frac{1}{\cos(x)} - 1}\ $$

Now I'm not sure if I'm allowed to use arithmetic of limits, as I don't know if the limit exist. I don't understand why I can't just say that: $$ \lim_{x\to 0} \sin(x) = 0 $$ and therefore the upper limit is also 0 (regardless of the denominator..

Thank you

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    $\begingroup$ why don't you try to use your idea on a function that satisfies $f(x)=\frac{x}{x}$ for $x\neq 0$? $\endgroup$ – HereToRelax Dec 29 '16 at 20:34
  • $\begingroup$ The denominator limits to 0 so the arithmetic of limits doesn't (immediately) apply. $\endgroup$ – Heath Winning Dec 29 '16 at 20:36
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    $\begingroup$ The reason why you cannot say that, is because even when the numerator tends to zero, if the denominator also tends to zero, and does so "just as fast or even faster", the denominator will "win" and the limit of the full expression could fail to be zero. Simple example: $$\lim_{x\to 0}\frac{x}{x^2}$$ $\endgroup$ – Jeppe Stig Nielsen Dec 30 '16 at 8:07
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HINT:

$$\frac{\sin(x)}{\frac1{\cos(x)}-1}=\frac{\frac12\sin(2x)}{1-\cos(x)}=\frac{\frac12\sin(2x)}{2\sin^2(x/2)}\sim \frac{x}{x^2/2}$$

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Write as $$\cos x\frac{\sin x}{x}\frac{x^2}{1-\cos x}\frac{1}{x}$$ All factors have a finite limit except the last, which as limit $\infty$, so it does not exist.

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    $\begingroup$ That last sentence needs work. "Finite" is not enough. You want to say "finite non-zero limit" Also, the limit is $-\infty$ from the left. $\endgroup$ – zhw. Dec 29 '16 at 20:56
  • $\begingroup$ I can't see why 1 - cos(x) isn't a problem.. cox(x) when x->0 is 1 so don't we need to keep simplifying it? $\endgroup$ – Noam Dec 29 '16 at 21:16
  • $\begingroup$ $\frac{1-\cos x}{x^2}\to\frac{1}{2}$ is a well known limit. $\endgroup$ – Rene Schipperus Dec 29 '16 at 21:27
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Hint:

$${\sin x\over{1\over\cos x}-1}={\sin x\cos x\over1-\cos x}={\sin x\cos x (1+\cos x)\over1-\cos^2x}={\cos x(1+\cos x)\over\sin x}$$

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  • $\begingroup$ Thanks for the comment but I still can't see it.. you have sinx in the denominator so you still can't use arithmetic of limits.. $\endgroup$ – Noam Dec 29 '16 at 21:13
  • $\begingroup$ @Noam, does Mark Bennet's answer make any more sense to you? Or Dr. MV's hint? We're all saying pretty much the same thing. $\endgroup$ – Barry Cipra Dec 30 '16 at 3:57
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    $\begingroup$ @Noam - note that this is no longer an indeterminant form. By Barry's work, the form is $\frac 2 0$. You can say something about that limit. If your limit converges, $\lim_{x\to 0}\frac{\sin x}{\frac{1}{\cos x} - 1} = L$, then $2 = \lim_{x\to 0} \cos x(1+\cos x) = \lim_{x\to 0} \sin x \frac{\sin x}{\frac{1}{\cos x} - 1} = 0 \cdot L = 0$. $\endgroup$ – Paul Sinclair Dec 30 '16 at 4:56
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Multiply through by $\cos x$ $(\text{NB }\neq 0)$ to clear fractions and obtain $$\frac {\sin x\cos x}{1-\cos x}=\frac {2\sin {\frac x2}\cos {\frac x2}\cos x}{2\sin^2 {\frac x2}}=\frac {\cos {\frac x2}\cos x}{\sin {\frac x2}}$$And now the behaviour should be obvious.

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Notice that

$$\lim_{x\to0}\frac xx=1$$

This should be fairly obvious since $x/x=1$.

But by your logic, this limit should be $0$.

But you forget that the denominator is also $0$, so it is actually an indeterminate form. If the denominator were anything else, then your reasoning would be right.

You should look towards the other answers to find more suitable methods for evaluating these types of limits.

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  • $\begingroup$ I understand, thank you. $\endgroup$ – Noam Dec 29 '16 at 20:40
  • $\begingroup$ @Noam No problem, just be careful with the way you handle limits. $\endgroup$ – Simply Beautiful Art Dec 29 '16 at 20:41

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