lim cos(sinx)- cosx/x4 x-0

limx→0cossinx-cosxx4It is 0/0 form so using L'hospital rule, limx→0ddxcossinx-cosxddxx4=limx→0-sinsinx×cosx+sinx4x3Again using same rule, =limx→0-cossinx×cos2x+sinxsinsinx+cosx12x2Again, =limx→0-cossinx×2cosx×-sinx+cos2xsinsinx×cosx+cosxsinsinx+sinxcossinx×cosx-sinx24xNow putting limit and using result limx→0sinxx=1 =2+1+1+1-124=16

lim cos(sinx)- cosx/x⁴
x-0

\lim_{x \to 0} \frac{\cos (sinx) - cosx}{x^{4}} It is 0 / 0 form so using L' hospital rule, \lim_{x \to 0} \frac{\frac{d}{dx} (\cos (sinx) - cosx)}{\frac{d}{dx} x^{4}} = \lim_{x \to 0} \frac{- \sin (sinx) \times cosx + sinx}{4 x^{3}} Again using same rule, = \lim_{x \to 0} \frac{- \cos (sinx) \times \cos^{2} x + sinxsin (sinx) + cosx}{12 x^{2}} Again, = \lim_{x \to 0} \frac{- \cos (sinx) \times 2 cosx \times - sinx + \cos^{2} xsin (sinx) \times cosx + cosxsin (sinx) + sinxcos (sinx) \times cosx - sinx}{24 x} Now putting limit and using result \lim_{x \to 0} \frac{sinx}{x} = 1 = \frac{2 + 1 + 1 + 1 - 1}{24} = \frac{1}{6}

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