Question
Download Solution PDFEvaluate \(\rm \displaystyle \lim_{x \rightarrow 0} \frac{1 - \cos 2x }{x^2} \)
Answer (Detailed Solution Below)
Detailed Solution
Download Solution PDFConcept:
\(\rm \displaystyle \lim_{x \rightarrow 0} \frac{\sin x }{x} =1\)
Trigonometry formulas:
1 - cos 2x = 2sin2 x
1 + cos 2x = 2cos2 x
Calculation:
Here, we have to find the value of the limit \(\rm \displaystyle \lim_{x \rightarrow 0} \frac{1 - \cos 2x }{x^2} \)
As we know, 1 - cos 2x = 2sin2 x
\(\rm \displaystyle \lim_{x \rightarrow 0} \frac{1 - \cos 2x }{x^2} \)
= \(\rm \displaystyle \lim_{x \rightarrow 0} \frac{2\sin^2 x }{x^2} \)
= 2 × \(\rm \displaystyle \lim_{x \rightarrow 0} \frac{\sin x }{x} × \lim_{x \rightarrow 0} \frac{\sin x }{x}\)
= 2 × 1 × 1
= 2
Last updated on Jun 30, 2025
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