TS EAMCET · Maths · Continuity and Differentiability
If \(f(x)=\lim _{n \rightarrow \infty}\left(\frac{\log (2+x)-x^{2 n} \sin x}{1+x^{2 n}}\right), 0 \leq x \leq \frac{\pi}{2}\) \(x \in \mathbf{R}\), then at \(x=1, f(x)\) is
- A differentiable
- B discontinuous
- C continuous
- D continuous but not differentiable
Answer & Solution
Correct Answer
(B) discontinuous
Step-by-step Solution
Detailed explanation
\begin{aligned} & \text { LHL. }=\lim _{x \rightarrow 1^{-}} \frac{\log (2+x)-x^{2 / 2} \sin x}{1+x^{2 n}} \\ & =\lim _{h \rightarrow 0} \frac{\log (2+1-h)-(1-h)^{2 n} \sin (1-h)}{1+(1-h)^{2 n}} \\ & \text { - } \lim _{h \rightarrow 10} \log (3-h) \\ & {\left[\because(1-h) 1…
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