If f(x) = lim_{n rightarrow infty} left( frac | vedclass

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(B) We analyze the limit $f(x) = \lim_{n \rightarrow \infty} \frac{\log(2+x) - x^{2n} \sin x}{1+x^{2n}}$.Case $1$: If $0 \leq x < 1$, then $x^{2n} \ri

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discontinuous

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(B) We analyze the limit $f(x) = \lim_{n ightarrow \infty} \frac{\log(2+x) - x^{2n} \sin x}{1+x^{2n}}$.Case $1$: If $0 \leq x Case $2$: If $x = 1$, then $f(1) = \lim_{n ightarrow \infty} \frac{\log(3) - 1^{2n} \sin(1)}{1+1^{2n}} = \frac{\log 3 - \sin 1}{2}$.Case $3$: If $1 Now, check continuity at $x=1$:Left-hand limit: $\lim_{x ightarrow 1^-} f(x) = \log(2+1) = \log 3$.Right-hand limit: $\lim_{x ightarrow 1^+} f(x) = -\sin(1)$.Since $\log 3 
eq -\sin 1$, the function is discontinuous at $x=1$.

If $f(x) = \lim_{n \rightarrow \infty} \left( \frac{\log(2+x) - x^{2n} \sin x}{1+x^{2n}} \right)$ for $0 \leq x \leq \frac{\pi}{2}$, then at $x=1$, $f(x)$ is

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