evaluate the limit as $x$ approaches $0$ of $(\cos x-1)/x$ without l'Hopital's rule

Question

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.

Answer 1

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*}

Answer 2

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 $.

Answer 3

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