Evaluating $\lim_{x\to0}\frac{1-\cos(x)}{x}$

$$\lim_{x\to0}\frac{1-\cos(x)}{x}$$

Could someone help me with this trigonometric limit? I'm trying to evaluate it without L'Hôpital's rule and derivation.

Solution 1:

Hint: Multiply top and bottom by $1+\cos x$. Then use a probably known fact about $\dfrac{\sin x}{x}$.

Solution 2:

$\displaystyle \lim_{x\to 0} \frac{1-\cos x}{x}= \lim_{x\to 0} \frac{(1-\cos x)(1+\cos x)}{ x(1+\cos x)}=\lim_{x\to 0} \displaystyle\frac{(1-\cos^2 x)}{ x(1+\cos x)}=\lim_{x\to 0} \frac{\sin^2 x}{ x(1+\cos x)}=\lim_{x\to 0} \frac{x\sin^2 x}{ x^2(1+\cos x)}=\lim_{x\to 0} \displaystyle\frac{\sin^2 x}{ x^2}\displaystyle\lim_{x\to 0}\frac{x}{1+\cos x}=1\times 0=0 $

O you simply apply L'Hopital rule and $$\displaystyle \lim_{x\to 0} \frac{1-\cos x}{x}=\displaystyle \lim_{x\to 0} \frac{(1-\cos x)'}{x'}=\displaystyle \lim_{x\to 0} \frac{\sin x}{1}=0$$

Solution 3:

As an alternative method, $\cos x=1+ O(x^2)$. Then $$O(x^2)/x= O(x)$$

Edit: The notation $f(x)=O(g(x))$ means that $f(x)/g(x)$ is bounded by a constant (in the limit). Put simply, this means $f$ is no bigger than $g$ in the limit.

This answer basically relies on the Taylor expansion of the numerator. Typically this is the quickest way to get to the answer with these sorts of limits.

For example, l'Hôpital's rule is seen to be because if $f(0)=g(0)=0$ for functions which can be approximated to some order by a Taylor series then $f(x)/g(x)= (x f'(0)+O(x^2))/(xg'(0)+O(x^2))$ and one can cancel the $x$s.

Solution 4:

$$\lim_{x\to0}\frac{1-\cos(x)}{x}$$ $$\lim_{x\to0}\frac{1-(1-2\sin^2 \dfrac x2)}{x}$$ $$\lim_{x\to0}\frac{2\sin^2\dfrac x2}{x}$$ $$\lim_{x\to0}\frac{2x\sin^2\dfrac x2}{4\left(\dfrac x2\right)^2}$$ $$\lim_{x\to0}\frac{x}{2}\cdot \lim_{x\to0}\left(\dfrac{\sin \dfrac x2}{\dfrac x2}\right)^2$$ $$0\cdot 1$$ $$0$$

Solution 5:

Using the definition of derivative: $$\lim\limits_{x \to 0} \frac{1-\cos(x)}{x}= - \lim\limits_{x \to 0} \frac{\cos(x)-\cos(0)}{x-0}=- \cos'(0)=-\sin(0)=0$$