11. find\n\n$$lim _ { \theta \rightarrow 0 } \frac { 1 - cos \theta } { sin \theta }$$
11. find\n\n$$lim _ { \theta \rightarrow 0 } \frac { 1 - cos \theta } { sin \theta }$$
Answer
Explanation:
Step1: Use trigonometric identities
Multiply numerator and denominator by (1 + \cos\theta). [ \begin{align*} \lim_{\theta\rightarrow0}\frac{1-\cos\theta}{\sin\theta}&=\lim_{\theta\rightarrow0}\frac{(1 - \cos\theta)(1+\cos\theta)}{\sin\theta(1+\cos\theta)}\ \end{align*} ] 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\rightarrow0}\frac{\sin^{2}\theta}{\sin\theta(1 + \cos\theta)})
Step2: Simplify the expression
Cancel out (\sin\theta) (for (\theta\neq0,\sin\theta\neq0)): (\lim_{\theta\rightarrow0}\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:
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