The limit of x sin left(e^{frac{1}{x}}right) | vedclass

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(A) We need to evaluate $\lim_{x ightarrow 0} x \sin \left(e^{\frac{1}{x}}ight)$.Since $-1 \leq \sin 	heta \leq 1$ for any real $	heta$, we have

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is equal to $0$

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(A) We need to evaluate $\lim_{x ightarrow 0} x \sin \left(e^{\frac{1}{x}}ight)$.Since $-1 \leq \sin 	heta \leq 1$ for any real $	heta$, we have $-1 \leq \sin \left(e^{\frac{1}{x}}ight) \leq 1$.Multiplying by $x$ (assuming $x > 0$): $-x \leq x \sin \left(e^{\frac{1}{x}}ight) \leq x$.As $x ightarrow 0^+$, both $-x$ and $x$ approach $0$. By the Squeeze Theorem, $\lim_{x ightarrow 0^+} x \sin \left(e^{\frac{1}{x}}ight) = 0$.For $x  0$. As $x ightarrow 0^-$, $h ightarrow 0^+$.The expression becomes $\lim_{h ightarrow 0^+} (-h) \sin \left(e^{-\frac{1}{h}}ight)$.As $h ightarrow 0^+$, $e^{-\frac{1}{h}} ightarrow e^{-\infty} = 0$, so $\sin \left(e^{-\frac{1}{h}}ight) ightarrow \sin(0) = 0$.Thus, $\lim_{h ightarrow 0^+} (-h) \cdot 0 = 0$.Since the left-hand limit and right-hand limit are both $0$, the limit exists and is equal to $0$.

The limit of $x \sin \left(e^{\frac{1}{x}}\right)$ as $x \rightarrow 0$

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