Question

1. find

$$lim _ { \theta ightarrow 0 } \frac { 1 - cos \theta } { sin \theta }$$

Explanation:

Step1: Use trigonometric identities

Multiply numerator and denominator by \(1 + \cos\theta\).

$$
LATEXBLOCK0
$$

Since \((a - b)(a + b)=a^{2}-b^{2}\), then \((1-\cos\theta)(1+\cos\theta)=1-\cos^{2}\theta=\sin^{2}\theta\). So the limit becomes \(\lim_{\theta
ightarrow0}\frac{\sin^{2}\theta}{\sin\theta(1 + \cos\theta)}\)

Step2: Simplify the expression

Cancel out \(\sin\theta\) (for \(\theta
eq0,\sin\theta
eq0\)): \(\lim_{\theta
ightarrow0}\frac{\sin\theta}{1+\cos\theta}\)

Step3: Substitute \(\theta = 0\)

Substitute \(\theta=0\) into \(\frac{\sin\theta}{1+\cos\theta}\). We know that \(\sin(0) = 0\) and \(\cos(0)=1\)

$$ \frac{\sin(0)}{1+\cos(0)}=\frac{0}{1 + 1}=0 $$

Answer:

\(0\)