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

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.

• 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 $\cos'(0)$. – runway44 Jun 30 at 0:11

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*}

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 $$\dfrac{\cos(x)-1}{x^2} \to -\dfrac12$$.

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