let\n$h(x)=\\begin{cases}e^{2x}&\\text{for }x < 0\\\\e^{5x}&\\text{for }x\\geq0\\end{cases}$\nis $h$…

Explanation:

Step1: Find left - hand limit

We find $\lim_{x\rightarrow0^{-}}h(x)$. Since for $x < 0$, $h(x)=e^{2x}$, then $\lim_{x\rightarrow0^{-}}h(x)=\lim_{x\rightarrow0^{-}}e^{2x}$. Substituting $x = 0$ into $e^{2x}$, we get $\lim_{x\rightarrow0^{-}}e^{2x}=e^{2\times0}=1$.

Step2: Find right - hand limit

We find $\lim_{x\rightarrow0^{+}}h(x)$. Since for $x\geq0$, $h(x)=e^{5x}$, then $\lim_{x\rightarrow0^{+}}h(x)=\lim_{x\rightarrow0^{+}}e^{5x}$. Substituting $x = 0$ into $e^{5x}$, we get $\lim_{x\rightarrow0^{+}}e^{5x}=e^{5\times0}=1$.

Step3: Find the function value at $x = 0$

Since for $x\geq0$, $h(x)=e^{5x}$, then $h(0)=e^{5\times0}=1$.

Step4: Check continuity condition

A function $y = h(x)$ is continuous at $x = a$ if $\lim_{x\rightarrow a^{-}}h(x)=\lim_{x\rightarrow a^{+}}h(x)=h(a)$. Here, $\lim_{x\rightarrow0^{-}}h(x)=\lim_{x\rightarrow0^{+}}h(x)=h(0) = 1$.

Answer:

A. Yes