Limits Continuity And Differentiability Question 80
Question: Let $ f(x)=\underset{n\to \infty }{\mathop{\lim }}\frac{\log (2+x)-x^{2n}\sin x}{1+x^{2n}} $ , then
Options:
A) f is continuous at x=1
B) $ \underset{x\to {1^{+}}}{\mathop{\lim }}f(x)=log3 $
C) $ \underset{x\to {1^{+}}}{\mathop{\lim }}f(x)=-\sin 1 $
D) $ \underset{x\to {1^{-}}}{\mathop{\lim }}f(x) $ does not exist
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Answer:
Correct Answer: C
Solution:
For $ | x |<1,x^{2n}\to 0 $ as $ n\to \infty $
and for $ | x |>1,1/x^{2n}\to 0 $ as $ n\to \infty , $ So, $ f(x)= \begin{vmatrix} \log (2+x) \\ \underset{\to \infty }{\mathop{\lim }}\frac{{x^{-2n}}\log (2+x)-sinx}{{x^{-2n}}+1}=-\sin x, \\ \frac{1}{2}[log(2+x)-sinx], \\ \end{vmatrix} .\begin{vmatrix} | x |<1 \\ | x |>1 \\ | x |=1 \\ \end{vmatrix} $
Thus, $ \underset{x\to {1^{+}}}{\mathop{\lim }}f(x)=\underset{x\to 1}{\mathop{\lim }}(-\sin x)=-sin1 $ and $ \underset{x\to {1^{-}}}{\mathop{\lim }}f(x)=\underset{x\to 1}{\mathop{\lim }}log(2+x)=\log 3. $
BETA
