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I am having trouble completing this proof.

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

The mean value theorem guarantees that we have a c such that $\displaystyle f'(c)= \frac{f(b)-f(a)}{b-a}$. In our case, we can write $\displaystyle \frac{f(x)-f(0)}{f-0}=f' (c)$. But that is just the definition of the derivative, so I don't see how to go from this to the limit?

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    $\begingroup$ Poor form to use the mean value theorem here. What you have is, by definition, the derivative of cosine, evaluated at $0$. $\endgroup$ – Andrés E. Caicedo Feb 25 '14 at 4:05
  • $\begingroup$ I am told to use the mean value theorem by the assignment. $\endgroup$ – clickclock Feb 25 '14 at 4:05
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    $\begingroup$ Which doesn't change the fact it is a bad question. $\endgroup$ – Andrés E. Caicedo Feb 25 '14 at 4:08
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Hint: Consider a very small positive $x$, together with $f(x) := \cos x$. Then

$$\frac{\cos x - 1}{x} = \frac{f(x) - f(0)}{x - 0} = f'(c) = \sin c$$

for some $0 < c < x$.

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  • $\begingroup$ If x is small, the numerator would be close to 1 and the denominator would be very small, so wouldn't the fraction blow up? $\endgroup$ – clickclock Feb 25 '14 at 4:07
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    $\begingroup$ The numerator goes to zero more quickly than the denominator does. $\endgroup$ – user61527 Feb 25 '14 at 4:41

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