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Prove that $$\lim_{x \to 0} \frac{\cos x -1}{x}=0$$ using the mean value theorem.

I understand that this may be shown with a different method, but I am required to use MVT. How can I prove this expression? I don't see how MVT is applicable here at all.

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Mean value theorem states that there is $c \in (a,b)$ such that: $$ f'(c) = \frac{f(b)-f(a)}{b-a} $$

Let $f(x) = \cos (x)\text{, } b=x \text{ and }a = 0$. This gives $$ - \sin(c) = \frac{\cos(x)-\cos(0)}{x-0}=\frac{\cos(x)-1}{x} $$ $$ x\rightarrow 0 \Rightarrow c\rightarrow0 \Rightarrow \sin(c) \rightarrow 0 $$

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  • $\begingroup$ Thank you! One question, how do we know that x->0 => c->0? $\endgroup$ – something Feb 26 '14 at 2:16
  • $\begingroup$ Remember that c has to be 0<c<x. $\endgroup$ – lamiaani Feb 26 '14 at 2:18

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