$\lim_{x\to 0} \frac{1-\cos(x)}{x\cos(x)} $ without L'hopital

Question

Please help me to solve this limit without using L'Hôpital's rule. I don't know what other method can't be used to solve this limit.

$$\lim_{x\to 0} \frac{1-\cos(x)}{x\cos(x)} $$

Answer 1

HINT:

$$\frac{1-\cos(x)}{x\cos(x)}=\frac{2\sin^2(x/2)}{x\cos(x)} \tag 1$$

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SPOILER ALERT: Scroll over the highlighted area to reveal the solution

Using $(1)$ we have $$\begin{align}\lim_{x\to 0}\frac{1-\cos(x)}{x\cos(x)}&=\lim_{x\to 0}\frac{2\sin^2(x/2)}{x\cos(x)}\\\\&=\left(\lim_{x\to 0}\frac{\sin(x/2)}{x/2}\right)\left(\lim_{x\to 0}\frac{\sin(x/2)}{\cos(x)}\right)\\\\&=(1)(0)\\\\&=0\end{align}$$

Answer 2

$\frac{1-\cos x}{x\cos x} = \frac{2 \sin^2 (x/2)}{x \cos x}= \frac{\sin (x/2)}{x/2} \frac{\sin (x/2)}{\cos x}$ The limit is $0$.

Answer 3

Use Taylor series

$$\cos(x) \approx 1 - \frac{x^2}{2}$$ thence

$$\frac{1 - \cos(x)}{x\cos(x)} = \frac{1 - \left(1 - \frac{x^2}{2}\right)}{x\cdot\left(1 - \frac{x^2}{2}\right)} = \frac{x^2}{2x - x^3} = \frac{x^2}{x^2\left(\frac{2}{x} - x\right)} = \frac{1}{\frac{2}{x}} = \frac{x}{2}$$

And since $x\to 0$ the limit is

$$\boxed{0}$$

Answer 4

This answer is a follow-up to André Nicolas' and Paul Sinclair's comments

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

Note that $\lim_{x\to 0}\frac{\sin x}{x}=1$, that $\lim_{x\to 0}\sin x=0$, that $\lim_{x\to 0}\cos x=1$, and that $\lim_{x\to 0}(1+\cos x)=2$ giving the final result $$\lim_{x\to 0}\frac{1-\cos x}{x\cos x}=0$$

Answer 5

Note that $ \displaystyle \lim_{x\to0}\dfrac{1-\cos x}{x} =-\lim_{x\to0}\dfrac{\cos x-\cos 0}{x - 0} =-\left.\cos'x\right|_{x=0} = \sin 0 = 0 $

So $ \displaystyle \lim_{x\to 0} \frac{1-\cos(x)}{x\cos(x)} = \lim_{x\to 0}\frac{1-\cos(x)}{x} \cdot \lim_{x\to 0}\dfrac{1}{\cos(x)} = 0 \cdot 1 = 0 $