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I'm very confused about this. When finding the derivative of sine, we have

$\lim_{h\to0}\dfrac{\sin(x+h)-\sin x}{h}=\dfrac{\lim_{h\to0}(\sin(x+h)-\sin x)}{\lim_{h\to0}h}=\dfrac{\lim_{h\to0}(\sin x\cos h+\sin h\cos x-\sin x)}{\lim_{h\to0}h}=\dfrac{\lim_{h\to0}(\sin x\cos h-x)+\lim_{h\to0}(\sin h\cos x)}{\lim_{h\to0}h}=\dfrac{0+\lim_{h\to0}(\sin h\cos x)}{\lim_{h\to0}h}=\cos x\lim_{h\to0}\dfrac{\sin h}{h}=\cos x$

Is this proof correct? If not, why is it correct to say $\lim_{x\to0}\dfrac1{x^2}=\dfrac1{0}=\infty$? Please explain.

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$\lim_{x\to0}\dfrac1{x^2}\ne\infty$, because the limit doesn't exist. You can only apply the limit laws when the limits exist. – rae306 Jul 6 '14 at 7:38
It does exist. – Klobbbyyy Jul 6 '14 at 7:43
Oops, mistake. Thanks! ;-) – rae306 Jul 6 '14 at 7:47
up vote 2 down vote accepted

Quotient rule can only be used when denominator's limit doesn't become zero.

$\lim_{h\to0}\dfrac{\sin(x+h)-\sin x}{h}\\ =\lim_{h\to0}\dfrac{(\sin(x+h)-\sin x)}{h}\\ =\lim_{h\to0}\dfrac{(\sin x\cos h+\sin h\cos x-\sin x)}{h}\\ =\lim_{h\to0}\dfrac{\sin x(\cos h-1)+(\sin h\cos x)}{h}\\ =\lim_{h\to0}\dfrac{(\cos h-1)}{h}.\sin x+\lim_{h\to0}\dfrac{\sin h}{h}.(\cos x)\\ =\cos x\lim_{h\to0}\dfrac{\sin h}{h}=\cos x $

If $\epsilon$ is any positive number, however small, then we can find a number $n_0$ such that $\frac1n < \epsilon$ for all values of n greater than or equal to $n_0$
We shall say that ‘the limit of $\frac1n$ as n tends to $\infty$ is $0$’, a statement which we may express symbolically in the form: $$\lim_{n\to\infty}\frac1n=0$$

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You factored incorrectly in line 4, it must be: $=\lim_{h\to0}\dfrac{\sin x(\cos h-1)+(\sin h\cos x)}{h}\\$ – rae306 Jul 6 '14 at 7:45
Thanks. There's a typo in the fourth line. – Klobbbyyy Jul 6 '14 at 7:46
how did you show $\lim_{h \to 0} \frac{\cos h -1}{h} = 0$? – Alex Jul 6 '14 at 10:00
@Alex That follows from $\lim_{h\to0}\frac{\sin h}{h}=1$. – Klobbbyyy Jul 6 '14 at 12:55

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