Question

h(x)=\

$$\begin{cases}\\sin(x)&\\text{for }x < 0\\\\\\sqrt{x + \\pi}&\\text{for }x\\geq0\\end{cases}$$

is h continuous at x = 0?
choose 1 answer:
a yes
b no

Explanation:

Step1: Find left - hand limit

We find $\lim_{x
ightarrow0^{-}}h(x)$. Since $x < 0$, $h(x)=\sin(x)$. So, $\lim_{x
ightarrow0^{-}}h(x)=\lim_{x
ightarrow0^{-}}\sin(x)=0$.

Step2: Find right - hand limit

We find $\lim_{x
ightarrow0^{+}}h(x)$. Since $x\geq0$, $h(x)=\sqrt{x + \pi}$. So, $\lim_{x
ightarrow0^{+}}h(x)=\lim_{x
ightarrow0^{+}}\sqrt{x+\pi}=\sqrt{0+\pi}=\sqrt{\pi}$.

Step3: Check continuity condition

For a function to be continuous at $x = a$, $\lim_{x
ightarrow a^{-}}h(x)=\lim_{x
ightarrow a^{+}}h(x)=h(a)$. Here, $\lim_{x
ightarrow0^{-}}h(x)=0$ and $\lim_{x
ightarrow0^{+}}h(x)=\sqrt{\pi}$, and $0
eq\sqrt{\pi}$.

Answer:

B. No