question_answer

Let\[f(x)=\underset{n\to \infty }{\mathop{\lim }}\,\frac{{{\log }_{e}}(2+x)-{{x}^{2n}}\sin x}{1+{{x}^{2n}}},\]then

A) f(x) is continuous at x = 1

B) \[\underset{x\to {{1}^{+}}}{\mathop{\lim }}\,f(x)=lo{{g}_{e}}3\]

C) \[\underset{x\to {{1}^{+}}}{\mathop{\lim }}\,f(x)=-\sin 1\]

D) \[\underset{x\to {{1}^{+}}}{\mathop{\lim }}\,f(x)\]does not exist

Correct Answer: C

Solution :

[c] : For \[|x|<1,{{x}^{2n}}\to 0\]as\[n\to \infty \] \[|x|>1,\frac{1}{{{x}^{2n}}}\to 0\]as\[n\to \infty \] \[\underset{x\to {{1}^{+}}}{\mathop{\lim }}\,(f)(x)=-sin1;\underset{x\to {{1}^{-}}}{\mathop{\lim }}\,f(x)=log3\]