1. Tardigrade
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  4. Let f (x)= displaystyle limn→∞ (log (2+x)-x2n sin x/1+x2n) Then

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Q. Let $ f \left(x\right)=\displaystyle\lim_{n\to\infty} \frac{log \left(2+x\right)-x^{2n}\,sin\,x}{1+x^{2n}}$ Then

Continuity and Differentiability

Solution:

For $\left|x\right|<\,1, x^{2n} \rightarrow 0$ as $n \rightarrow \infty$, and for $\left|x\right|>\,1, 1 /x^{2n}\rightarrow 0$ as
$n \rightarrow\infty $ So,
$f(x) = \begin{cases} log(2+x), & \quad \text{ } |x|<\,1 \\ \displaystyle\lim_{n\to\infty} \frac{x^{-2n} log\left(2+x\right)-sin\,x}{x^{-2n}+1}=-sin\,x, & \quad \text{ } |x|>\,1 \\ \frac{1}{2}[log (2+x)-sin\,x], & \quad \text{ } |x|=1 \end{cases} $
Thus,
$\displaystyle\lim_{x\to1^{+}} f \left(x\right) =\displaystyle\lim_{x\to1}\left(-sin\,x\right)=-sin\,1$
and
$\displaystyle\lim_{x\to1^{-}} f \left(x\right)=\displaystyle\lim_{x\to1} log\left(2+x\right)=log\,3$