Question
$\lim _{x \rightarrow 2} \frac{\sin \left(e^{x-2}-1\right)}{\ln (x-1)}$ is equal to(a) $-2$(b) $-1$(c) 0(d) 1
Step 1
As $x \rightarrow 2$, we have $u \rightarrow e^0 - 1 = 0$. So, we can rewrite the limit as: $\lim_{u \rightarrow 0} \frac{\sin u}{\ln (e^u)}$ Show more…
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