The value of the limit lim_{x rightarrow 1} f | vedclass

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(D) Let $x = 1 + h$. As $x \rightarrow 1$, $h \rightarrow 0$. Substituting this into the limit: $\lim_{h \rightarrow 0} \frac{\sin(e^{(1+h)-1}-1)}{\lo

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$1$

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(D) Let $x = 1 + h$. As $x \rightarrow 1$, $h \rightarrow 0$. Substituting this into the limit: $\lim_{h \rightarrow 0} \frac{\sin(e^{(1+h)-1}-1)}{\log(1+h)} = \lim_{h \rightarrow 0} \frac{\sin(e^h-1)}{\log(1+h)}$ We know that $\lim_{u \rightarrow 0} \frac{\sin u}{u} = 1$, $\lim_{h \rightarrow 0} \frac{\log(1+h)}{h} = 1$, and $\lim_{h \rightarrow 0} \frac{e^h-1}{h} = 1$. Rewriting the expression: $\lim_{h \rightarrow 0} \left( \frac{\sin(e^h-1)}{e^h-1} \cdot \frac{e^h-1}{h} \cdot \frac{h}{\log(1+h)} \right)$ $= 1 \cdot 1 \cdot \frac{1}{1} = 1$.

The value of the limit $\lim_{x \rightarrow 1} \frac{\sin(e^{x-1}-1)}{\log x}$ is

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