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JEE MainMathematicsModerateLimits

If
 lim⁡x→0\mspace2mu1−cos⁡3xxsinx⋅cosx \ \lim_{x \rightarrow 0}\mspace{2mu}\frac{1 - \cos^{3}x}{xsinx \cdot cosx}\ is
equal to 
λ\lambda then the value of 4λ4\lambda is

Show answer and solution
lim⁡x→0\mspace2mu1−cos⁡3xxsinx⋅cosx=lim⁡x→0\mspace2mu(1−cosx)⋅(1+cos⁡2x+cosx)x2⋅(sinxx)⋅cosx\lim_{x \rightarrow 0}\mspace{2mu}\frac{1 - \cos^{3}x}{xsinx \cdot cosx} = \lim_{x \rightarrow 0}\mspace{2mu}\frac{(1 - cosx) \cdot \left( 1 + \cos^{2}x + cosx \right)}{x^{2} \cdot \left( \frac{sinx}{x} \right) \cdot cosx}
=lim⁡x→0\mspace2mu(1−cos⁡xx2)⋅lim⁡x→0\mspace2mu1(sin⁡xx)⋅lim⁡x→0\mspace2mu(1+cos⁡2x+cos⁡xcos⁡x)= \lim_{x \rightarrow 0}\mspace{2mu}\left( \frac{1 - \cos x}{x^{2}} \right) \cdot \lim_{x \rightarrow 0}\mspace{2mu}\frac{1}{\left( \frac{\sin x}{x} \right)} \cdot \lim_{x \rightarrow 0}\mspace{2mu}\left( \frac{1 + \cos^{2}x + \cos x}{\cos x} \right)
=(12)⋅(1)⋅(31)=32= \left( \frac{1}{2} \right) \cdot (1) \cdot \left( \frac{3}{1} \right) = \frac{3}{2}

Source: Grass-G / M-19 Mathematics Paper 28 Nov Fourth.docx

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