If f(x) = frac{sin(e^{x-2} - 1)}{log(x-1)}, t | vedclass

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(C) Given $\lim_{x 	o 2} f(x) = \lim_{x 	o 2} \frac{\sin(e^{x-2} - 1)}{\log(x-1)}$.Let $t = x - 2$. As $x 	o 2$,$t 	o 0$. Then $x = t + 2$.Substit

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$1$

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(C) Given $\lim_{x 	o 2} f(x) = \lim_{x 	o 2} \frac{\sin(e^{x-2} - 1)}{\log(x-1)}$.Let $t = x - 2$. As $x 	o 2$,$t 	o 0$. Then $x = t + 2$.Substituting this into the limit:$\lim_{t 	o 0} \frac{\sin(e^t - 1)}{\log(t + 2 - 1)} = \lim_{t 	o 0} \frac{\sin(e^t - 1)}{\log(1 + t)}$.We know that $\lim_{u 	o 0} \frac{\sin u}{u} = 1$ and $\lim_{t 	o 0} \frac{\log(1 + t)}{t} = 1$.Multiplying and dividing by $(e^t - 1)$ and $t$:$\lim_{t 	o 0} \left( \frac{\sin(e^t - 1)}{e^t - 1} ight) 	imes \left( \frac{e^t - 1}{t} ight) 	imes \left( \frac{t}{\log(1 + t)} ight)$.Since $\lim_{t 	o 0} (e^t - 1) = 0$,the first term approaches $1$.$\lim_{t 	o 0} \frac{e^t - 1}{t} = 1$ and $\lim_{t 	o 0} \frac{t}{\log(1 + t)} = 1$.Therefore,the limit is $1 	imes 1 	imes 1 = 1$.

If $f(x) = \frac{\sin(e^{x-2} - 1)}{\log(x-1)},$ then $\lim_{x \to 2} f(x)$ is given by

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