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How can I find the following problems using elementary trigonometry?

$$\lim_{x\to 0}\frac{1−\cos x}{x^2}.$$

$$\lim_{x\to0}\frac{\tan x−\sin x}{x^3}. $$

Have attempted trig identities, didn't help.

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  • $\begingroup$ Am I correct in assuming that you either do not know or are not allowed to use L'Hospital's rules? $\endgroup$ – 5xum Jun 2 '14 at 11:12
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    $\begingroup$ Taylor series saves you every time. If you don't like that, you can always use half-angle formulas. The first numerator is $2\sin^2 \frac{x}{2}$ and the second is very similar once you write it as $\tan x (1-\cos x)$. $\endgroup$ – orion Jun 2 '14 at 11:12
  • $\begingroup$ I am allowed to use L'Hospital's rule! $\endgroup$ – user152739 Jun 2 '14 at 11:40
  • $\begingroup$ @user152739 Then just use the rule a couple of times, it's simple. $\endgroup$ – 5xum Jun 2 '14 at 11:46
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Using $\cos x = 1-2\sin^2\frac x2$ gives you $$\frac{1-\cos x}{x^2} = \frac{2\sin^2\frac x2}{x^2} = \frac12\left(\frac{\sin\frac x2}{\frac x2}\right)^2.$$

Using similar identities will help with the second limit as well.

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HINT : \begin{align} \lim_{x\to 0}\frac{1−\cos x}{x^2}&=\lim_{x\to 0}\frac{1−\cos x}{x^2}\cdot\frac{1+\cos x}{1+\cos x}\\ &=\lim_{x\to 0}\frac{1−\cos^2 x}{x^2(1+\cos x)}\\ &=\lim_{x\to 0}\frac{\sin^2 x}{x^2(1+\cos x)}\\ &=\lim_{x\to 0}\frac{\sin x\cdot\sin x}{x\cdot x\cdot(1+\cos x)}\\ &=\lim_{x\to 0}\frac{\sin x}{x}\cdot\lim_{x\to 0}\frac{\sin x}{x}\cdot\lim_{x\to 0}\frac{1}{1+\cos x}\\ &=1\cdot1\cdot\frac1{1+1} \end{align} and \begin{align} \lim_{x\to 0}\frac{\tan x−\sin x}{x^3}&=\lim_{x\to 0}\frac{\tan x−\sin x}{x^3}\cdot\frac{\cos x}{\cos x}\\ &=\lim_{x\to 0}\frac{\sin x(1−\cos x)}{x^3\cos x}\\ &=\lim_{x\to 0}\frac{\sin x(1−\cos x)}{x^3\cos x}\cdot\frac{1+\cos x}{1+\cos x}\\ &=\lim_{x\to 0}\frac{\sin^3 x}{x^3\cos x(1+\cos x)},\\ \end{align} where $\displaystyle\lim_{x\to 0}\frac{\sin x}{x}=1.$

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  • $\begingroup$ Should I take the derivative then? I can't seem to get anywhere with it! $\endgroup$ – user152739 Jun 2 '14 at 11:42
  • $\begingroup$ @user152739 No. See my edit for the first one and use the similar way to the second one. $\endgroup$ – Tunk-Fey Jun 2 '14 at 11:50
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L'hopital's rule is useful in this case, since the limits are of indeterminate form $0/0$. Then, taking the derivative of the numerator and denominator: $$ \lim_{x\to 0}\frac{1−\cos x}{x^2} = \lim_{x\to 0}\frac{\sin x}{2x} = \frac{1}{2}\lim_{x\to 0} \frac{\sin x}{x}. $$ which should look more familiar. The second limit is the same deal, except you apply L'hopital's rule 3 times (until the denominator is a constant). See if you can work it out yourself!

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