\(\rm \displaystyle \lim_{\theta\rightarrow 0}\frac{1-\cos m \theta}{1-\cos n\theta}=?\)

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  1. \(\rm \frac{m}{n}\)
  2. \(\rm \frac{m^2}{n}\)
  3. \(\rm \frac{m}{n^2}\)
  4. \(\rm \frac{m^2}{n^2}\)

Answer (Detailed Solution Below)

Option 4 : \(\rm \frac{m^2}{n^2}\)
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ఇచ్చినది:

\(\rm \displaystyle \lim_{θ\rightarrow 0}\frac{1-\cos m θ}{1-\cos nθ}\)

భావన:

పరిమితి భావనను ఉపయోగించండి.

\(\rm \lim_{x\rightarrow0}\frac{sinx}{x}=1\)

మరియు సూత్రం \(\rm cos2θ=1-2sin^2θ\)

గణన:

\(\rm \displaystyle \lim_{θ\rightarrow 0}\frac{1-\cos m θ}{1-\cos nθ}\)

\(\rm \displaystyle =\lim_{θ\rightarrow 0}\frac{1-(1-2\sin^2 \frac{m θ}{2})}{1-(1-2\sin^2 \frac{nθ}{2})}\)

\(\rm \displaystyle =\lim_{θ\rightarrow 0}\frac{2\sin^2 \frac{m θ}{2}}{2\sin^2 \frac{nθ}{2}}\)

\(\rm \displaystyle =\frac{\lim_{θ\rightarrow 0}\sin^2 \frac{m θ}{2}}{\lim_{θ\rightarrow 0}\sin^2 \frac{nθ}{2}}\)

m2θ/n2θ తో గుణించి భాగించండి

\(\rm =\frac{m^2θ}{n^2θ}=\frac{m^2}{n^2}\)

కాబట్టి 4వ ఎంపిక సరైనది.

 

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