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$$\lim_{x \to 0} \frac{1-\cos2x}{x\sin(x+4\pi)}$$ .... and then I got ... $$= \lim_{x \to 0} \frac{2\sin^2x}{x \sin x}$$ and I can't insert 0 in that calculation as long as there's x in the denominator and I don't know how to "get rid of it".

Any ideas?

$\sin4\pi=0$ and $\cos4\pi=1$, is that correct?

Martin Sleziak
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L_McClain
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  • so if I use $$= \lim_{x \to 0} \frac{sinx}{x}=1$$ i get $$= \lim_{x \to 0} \frac{2sinx}{sinx}=$$ and then a slution is 2? – L_McClain Dec 01 '13 at 14:19
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    Knowing this limit should be useful here: http://math.stackexchange.com/questions/75130/how-to-prove-that-lim-limits-x-to0-frac-sin-xx-1 – Martin Sleziak Dec 01 '13 at 14:19

3 Answers3

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Just continue start by dividing both fraction sides by $\sin x$: $$\lim_{x \to 0} \frac{2\sin^2x}{x\sin x}=\lim_{x \to 0} \frac{2\sin x}{x}=2\lim_{x \to 0} \frac{\sin x}{x}=2$$

Martin Sleziak
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LeeNeverGup
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Remember $\displaystyle x\to0\implies$

$\displaystyle(i) x\ne0$

$\displaystyle(ii)\sin x\to0\implies\sin x\ne0$

So,we can write $\displaystyle\lim_{h\to0}\frac hh=1$ and we know $\displaystyle\lim_{h\to0}\frac{\sin h}h=1$

lab bhattacharjee
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You can simplify the fraction first by dividing the top and the bottom by $\sin x$

$$ \lim_{x \to 0} \frac{2\sin^2x}{x\sin x} = \lim_{x \to 0} \frac{2\sin x}{x}$$

$\lim_{x \to 0} \frac{\sin x}{x}$ = 1, so: $$\lim_{x \to 0} \frac{2\sin x}{x} = 2$$

user93089
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