what is the lim_{x→ln2} g(x) if g(x) = {e^x if x > ln2, 4 - e^x if x ≤ ln2}? (a) 4 (b) 2 (c) e^2 (d) 0 (e) ln2

Explanation:

Step1: Recall the definition of limit

We need to find $\lim_{x\rightarrow\ln 2}g(x)$ where $g(x)=\begin{cases}4 - e^{x},&x\leq\ln 2\e^{x},&x > \ln 2\end{cases}$. We will find the left - hand limit and the right - hand limit.

Step2: Calculate the left - hand limit

For the left - hand limit as $x\rightarrow\ln 2^{-}$, we use the formula $g(x)=4 - e^{x}$. Substitute $x = \ln 2$ into $4 - e^{x}$. Since $e^{\ln 2}=2$, then $\lim_{x\rightarrow\ln 2^{-}}g(x)=4 - e^{\ln 2}=4 - 2=2$.

Step3: Calculate the right - hand limit

For the right - hand limit as $x\rightarrow\ln 2^{+}$, we use the formula $g(x)=e^{x}$. Substitute $x=\ln 2$ into $e^{x}$. Since $e^{\ln 2}=2$, then $\lim_{x\rightarrow\ln 2^{+}}g(x)=e^{\ln 2}=2$.

Step4: Determine the limit

Since $\lim_{x\rightarrow\ln 2^{-}}g(x)=\lim_{x\rightarrow\ln 2^{+}}g(x)=2$, then $\lim_{x\rightarrow\ln 2}g(x)=2$.

Answer:

B. 2