Differentiation Question 137
Question: Let $ f(x)=\underset{n\to \infty }{\mathop{\lim }}\frac{\log (2+x)-x^{2n}\sin x}{1+x^{2n}} $ . Then
Options:
A) $ \underset{x\to {1^{+}}}{\mathop{\lim }}f(x)\ne \underset{x\to {1^{-}}}{\mathop{\lim }}f(x) $
B) $ \underset{x\to {1^{+}}}{\mathop{\lim }}f(x)=sin1 $
C) $ \underset{x\to {1^{-}}}{\mathop{\lim }}f(x) $ doesn-t exist
D) None of these
Show Answer
Answer:
Correct Answer: A
Solution:
[a] For $ | x |<1,x^{2n}\to 0 $ as $ n\to \infty $ and For $ | x |>1,\frac{1}{x^{2n}}\to 0 $ as $ n\to \infty $ .
So, $ f(x)= \begin{cases} \log (2+x)if| x |<1 \\ \underset{n\to \infty }{\mathop{\lim }}\frac{{x^{-2n}}\log (2+x)-sinx}{{x^{-2n}}+1}=-\sin xif| x |>1 \\ \frac{1}{2}[log(2+x)-sinx]if| x |=1 \\ \end{cases} . $
Thus $ \underset{x\to {1^{+}}}{\mathop{\lim }}f(x)=\underset{x{{\to }^{1+}}}{\mathop{\lim }}(-sinx)=-sin1 $
and $ \underset{x\to {1^{-}}}{\mathop{\lim }}f(x)=\underset{x{{\to }^{1-}}}{\mathop{\lim }}\log (2+x)=log3. $
Hence, L.H.L $ \ne $ R.H.L
BETA
