Limit of function involving sin and cos

Question

How does one calculate the limit of...

$\frac{(1-cos(x))cos(x)}{sin(x)}$

as $x \rightarrow 0^+$

We haven't had much to do with trigonometric identities, so I probably am not supposed to use those to solve this one.

Answer 1

$$\frac{\cos x(1-\cos x)}{\sin x}=\frac{\cos x(1-\cos^2x)}{\sin x(1+\cos x)}=\frac{\sin x\cos x}{1+\cos x}\xrightarrow[x\to 0]{}\frac02=0$$

Answer 2

HINT:

As $1-\cos^2x=\sin^2x,$

$$\frac{1-\cos x}{\sin x}=\frac{\sin x}{1+\cos x}$$

Answer 3

Hint: $$\lim_{x\to0^+}\frac{(1-\cos x)\cos x}{\sin x }=\lim_{x\to0^+}\dfrac{1-\cos x}{x}\cdot\dfrac{x}{\sin x}\cdot\cos x.$$ You should recognize some known limits.

Answer 4

$$\frac{\cos x(1-\cos x)}{\sin x}$$

Multiply top and bottom by $(1+cos x)$ $$ \frac{\cos x(1-\cos^2x)}{\sin x(1+\cos x)}$$

You have to use the trignometric identity $sin^2x+cos^2x=1$ , Rearranging gives $1-cos^2x=sin^2x$

$$\frac{\sin x\cos x}{1+\cos x} $$

When

$$x\to 0\frac{\sin x\cos x}{1+\cos x}=\frac02=0$$