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I know with l'Hopital's rule it becomes $-\sin(x)$ which has the limit $0$.
However, I have been wondering how to evaluate this limit without l'Hopital's rule.

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  • $\begingroup$ Are you able to use the fact that the derivative of $\cos x$ is $-\sin x$? Because that limit is literally the limit definition of the derivative $\cos'(0)$. $\endgroup$ – runway44 Jun 30 '19 at 0:11
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HINT \begin{align*} \lim_{x\rightarrow 0}\frac{\cos(x)-1}{x} & = \lim_{x\rightarrow 0}\frac{(\cos(x)-1)(\cos(x)+1)}{x(\cos(x)+1)}\\\\ & = \lim_{x\rightarrow 0}\frac{\cos^{2}(x)-1}{x(\cos(x)+1)} = -\lim_{x\rightarrow 0}\frac{\sin^{2}(x)}{x(\cos(x)+1)} \end{align*}

Then make use of the fundamental limit \begin{align*} \lim_{x\rightarrow 0}\frac{\sin(x)}{x} = 1 \end{align*}

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It is hard to do without using $\dfrac{\sin(x)}{x} \to 1 $ so I won't try.

Here is one way using that.

$1-\cos(x) =2\sin^2(x/2) $ so

$\begin{array}\\ \dfrac{\cos(x)-1}{x} &=\dfrac{-2\sin^2(x/2)}{x}\\ &=\dfrac{-\sin^2(x/2)}{x/2}\\ &=-\dfrac{\sin^2(x/2)}{(x/2)^2}(\dfrac{x}{2})\\ &\to -\dfrac{x}{2}\\ &\to 0\\ \end{array} $

Note that this also shows the more precise result $\dfrac{\cos(x)-1}{x^2} \to -\dfrac12 $.

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$\cos x =1-\dfrac{x^{2}}{2}+...=1+o(x)$. Hence $\frac {\cos\,x -1} x \to 0$ as $x \to 0$,

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