Lim x - 0cos x-a sin b x1-x

Lim_x → 0 (cos x + a sin bx )^1/x = lim_x → (1 + (cos x + a sin bx 1)) ^1/x = e^lim_x → 0(cos x + a sin bx 1)/x = e^lim_x → 0(a sin bx (1 cos x ))/x ...

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Byju's Answer

Standard XII

Mathematics

Indeterminate Forms

lim x → 0cos ...

Question

${lim}_{x \to 0} (cos x + a sin b x)^{\frac{1}{x}}$

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Solution

${lim}_{x \to 0} (cos x + a sin b x)^{\frac{1}{x}} = {lim}_{x \to} (1 + (cos x + a sin b x - 1))^{\frac{1}{x}} = e^{{lim}_{x \to 0}} \frac{(cos x + a sin b x - 1)}{x} = e^{{lim}_{x \to 0}} \frac{(a sin b x - (1 - cos x))}{x} = e^{{lim}_{x \to 0}} \frac{(a sin b x - 2 {sin}^{2} (\frac{x}{2}))}{x} = e^{{lim}_{x \to} \frac{a b sin b x}{b x} - {lim}_{x \to 0} \frac{2 sin (\frac{x}{2}) sin (\frac{x}{2})}{2 (\frac{x}{2})}} = e^{{lim}_{x \to 0} \frac{a b sin b x}{b x} - {lim}_{x \to 0} \frac{2 sin (\frac{x}{2}) sin (\frac{x}{2})}{(\frac{x}{2})}} = e^{a b - 0} = e^{a b}$

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limx→0(cosx+asinbx)1x

limx→0(cosx+asinbx)1x=limx→(1+(cosx+asinbx−1))1x=elimx→0(cosx+asinbx−1)x=elimx→0(asinbx−(1−cosx))x=elimx→0(asinbx−2sin2(x2))x=elimx→absinbxbx−limx→02sin(x2)sin(x2)2(x2)=elimx→0absinbxbx−limx→02sin(x2)sin(x2)(x2)=eab−0=eab

${lim}_{x \to 0} (cos x + a sin b x)^{\frac{1}{x}}$