Limit Problem involving L'Hopital's Rule

Since plugging in $\ln(2)$ here gives us a $\cfrac{0}{0}$ indeterminate form, we can use L'Hopital's Rule and differentiate the top and bottom: $\begin{aligned} \lim\limits_{x \to \ln(2)} \frac{e^x - 2}{x-\ln(2)} \implies \lim\limits_{x \to \ln(2)} \frac{\frac{d}{dx}(e^x - 2)}{\frac{d}{dx}(x-\ln(2))} &= \lim\limits_{x \to \ln(2)} \frac{e^x}{1} \\ &= \lim\limits_{x \to \ln(2)} e^x \\ &= \green{\boxed{2}} \end{aligned}$

$\begin{aligned} L & = \lim_{x \to \ln 2} \frac {e^x - 2}{x - \ln 2} & \small \blue{\text{A 0/0 case, L'Hôpital's rule applies}} \\ & = \lim_{x \to \ln 2} \frac {e^x}1 & \small \blue{\text{Differentiate up and down w.r.t. }x} \\ & = e^{\ln 2} = \boxed 2 \end{aligned}$

Reference: L'Hôpital's rule