$\lim_{x\to 0} \frac{1-\cos(x)}{x\cos(x)} $ without L'hopital 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.
$$\lim_{x\to 0} \frac{1-\cos(x)}{x\cos(x)}  $$
 A: 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
$$\lim_{x\to 0}\frac{1-\cos x}{x\cos x}=0$$
A: 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
$$\boxed{0}$$
A: 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
$
A: $\frac{1-\cos x}{x\cos x} = \frac{2 \sin^2 (x/2)}{x \cos x}= \frac{\sin (x/2)}{x/2} \frac{\sin (x/2)}{\cos x}$
The limit is $0$. 
A: HINT:
$$\frac{1-\cos(x)}{x\cos(x)}=\frac{2\sin^2(x/2)}{x\cos(x)} \tag 1$$

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

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

A: frist :
$$\lim_{x \to 0} \frac{1-\cos x}{x^2}=\lim_{x \to 0} \frac{2\sin^2\frac{x}{2}}{x^2}=\frac{1}{2}$$
now :
$$ \lim_{x \to 0} \frac{1-\cos x}{x\cos x}=\lim_{x \to 0} \frac{1-\cos x}{x^2}.\frac{x^2}{x\cos x}\\=\lim_{x \to 0} \frac{1-\cos x}{x^2}.\frac{x}{\cos x}=?$$
since:
$$\lim_{x \to 0} \frac{x}{\cos x}=0$$
so :
$$\lim_{x \to 0} \frac{1-\cos x}{x^2}.\frac{x}{\cos x}=(\frac{1}{2})(0)=0$$
