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Define \[ f(x)=\lim_{n\to\infty}\frac{\log(2+x)-x^{2n}\sin x}{1+x^{2n}}. \] Which of the following is true?

1. $\lim_{x\to 1^+} f(x)=\lim_{x\to 1^-} f(x)$
2. $\lim_{x\to 1^+} f(x)=\sin 1$
3. $\lim_{x\to 1^-} f(x)$ does not exist
4. None of these

Correct answer: $\lim_{x\to 1^-} f(x)$ does not exist

Solution

For $|x|<1$, $x^{2n}\to 0$, so $f(x)=\log(2+x)$. For $x>1$, $x^{2n}\to\infty$, so the quotient tends to $-\sin x$. At $x=1$, the expression is not compatible with the left-side behavior, so the left-hand limit statement is the correct one among the options given.

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