KCET · Maths · Probability
\( \lim _{x \rightarrow 0} \frac{1-\cos x}{x^{2}} \) is
- A \( 02 \)
- B \( 03 \)
- C \( \frac{1}{2} \)
- D \( \frac{1}{3} \)
Answer & Solution
Correct Answer
(C) \( \frac{1}{2} \)
Step-by-step Solution
Detailed explanation
Given that \( \lim _{x \rightarrow 0} \frac{1-\cos x}{x^{2}} \)
It is \( \left(\frac{0}{0}\right) \) form
So, applying L'Hospitals rule, we have
\( \lim _{x \rightarrow 0} \frac{0-(0-\sin x)}{2 x}=\lim _{x \rightarrow 0} \frac{\sin x}{2 x} \)
It is again \( \left(\frac{0}{0}\right) \) form.
So, again applying L'Hospitals rule, we have
\( \lim _{x \rightarrow 0} \frac{\cos x}{2}=\frac{1}{2} \)
It is \( \left(\frac{0}{0}\right) \) form
So, applying L'Hospitals rule, we have
\( \lim _{x \rightarrow 0} \frac{0-(0-\sin x)}{2 x}=\lim _{x \rightarrow 0} \frac{\sin x}{2 x} \)
It is again \( \left(\frac{0}{0}\right) \) form.
So, again applying L'Hospitals rule, we have
\( \lim _{x \rightarrow 0} \frac{\cos x}{2}=\frac{1}{2} \)
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