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Limit x Tends to 0 1-cos x√cos2x/x^2 |$\lim_{x\rightarrow 0}\frac{1 – cosx\sqrt{\cos 2x}}{x^2}$

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Limit x Tends to 0 1-cos x√cos2x/x^2 |$\lim_{x\rightarrow 0}\frac{1 – cosx\sqrt{\cos 2x}}{x^2}$-This is an Important Math Problem of limit Chapter of Class 11 of Different Important Books Like RD Sharma, SN Day, NCERT, CBSE, WBCHSE, JEE, Etc.

Limit x Tends to 0 1-cos x√cos2x/x^2

$\lim_{x\rightarrow 0}\frac{1 – cosx\sqrt{\cos 2x}}{x^2}$
= $\lim_{x\rightarrow 0}\frac{(1 – cosx\sqrt{\cos 2x})(1 + cosx\sqrt{\cos 2x})}{x^2(1 + cosx\sqrt{\cos 2x})}$
= $\lim_{x\rightarrow 0}\frac{1 – \left(cosx\sqrt{\cos 2x}\right)^2}{x^2(1 + cosx\sqrt{\cos 2x})}$
= $\lim_{x\rightarrow 0}\frac{1 – cos^2x\ldotp cos2x}{x^2(1 + cosx\sqrt{\cos 2x})}$
= $\lim_{x\rightarrow 0}\frac{1 – cos^2x\left(1 – 2sin^2x\right)}{x^2(1 + cosx\sqrt{\cos 2x})}$
= $\lim_{x\rightarrow 0}\frac{1 – cos^2x + 2sin^2xcos^2x}{x^2(1 + cosx\sqrt{\cos 2x})}$
= $\lim_{x\rightarrow 0}\frac{\sin^2x + cos^2x – cos^2x + 2sin^2xcos^2x}{x^2(1 + cosx\sqrt{\cos 2x})}$$[\because \sin^2x + cos^2x = 1]$
= $\lim_{x\rightarrow 0}\frac{\sin^2x(1 + 2cos^2x)}{x^2(1 + cosx\sqrt{\cos 2x})} $
= $\lim_{x\rightarrow 0}\frac{\sin^2x}{x^2}\ldotp \lim_{x\rightarrow 0}\frac{(1 + 2cos^2x)}{(1 + cosx\sqrt{\cos 2x})} $
= $\lim_{x\rightarrow 0}\left(\frac{\sin x}{x}\right)^2\ldotp \frac{1 + 2}{1 + 1}$[$\because cos0°$ = 1]
= $\frac{3}{2}\left(\lim_{x\rightarrow 0}\frac{\sin x}{x}\right)^2 $
= $\frac{3}{2}$

আরও দেখুন

  1. Limit x Tends to Zero $\lim_{x\rightarrow 0}\frac{1 – cosx}{x^2}$
  2. Limit x Tends to Zero $\lim_{x\rightarrow 0}\frac{1 – cosx}{\sin^2x}$
  3. Limit x Tends to Zero $\lim_{x\rightarrow 0}\frac{xtanx}{1 – cosx}$
  4. Limit x Tends to 0 (tan x-sin x)/(1-cos x)|$\lim_{x\rightarrow 0}\frac{\tan x – sinx}{1 – cosx}$
  5. Evaluate: Limit x tends to zero (cosx-secx)/x^2||$\lim_{x\rightarrow 0}\frac{\cos x – secx}{x^2}$

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