g(x)=\\begin{cases}5 - x&\\text{for }x\\leq3\\\\2e^{3 - x}&\\text{for }x > 3\\end{cases}\nis g continuous at…

Explanation:

Step1: Calculate left - hand limit

We find $\lim_{x\rightarrow3^{-}}g(x)$. Since $x\rightarrow3^{-}$ means $x < 3$, we use $g(x)=5 - x$. So, $\lim_{x\rightarrow3^{-}}g(x)=\lim_{x\rightarrow3^{-}}(5 - x)=5-3 = 2$.

Step2: Calculate right - hand limit

We find $\lim_{x\rightarrow3^{+}}g(x)$. Since $x\rightarrow3^{+}$ means $x>3$, we use $g(x)=2e^{3 - x}$. Then $\lim_{x\rightarrow3^{+}}g(x)=\lim_{x\rightarrow3^{+}}2e^{3 - x}=2e^{3 - 3}=2e^{0}=2$.

Step3: Calculate function value

We find $g(3)$. Since $x = 3$, we use $g(x)=5 - x$, so $g(3)=5 - 3=2$.

Step4: Check continuity condition

A function $y = g(x)$ is continuous at $x=a$ if $\lim_{x\rightarrow a^{-}}g(x)=\lim_{x\rightarrow a^{+}}g(x)=g(a)$. Here, $\lim_{x\rightarrow3^{-}}g(x)=\lim_{x\rightarrow3^{+}}g(x)=g(3) = 2$.

Answer:

A. Yes