Question

g(x) = { (x - 2)^2 for x <= 2; 2 - x^2 for x > 2. is g continuous at x = 2? choose 1 answer: a yes b no

Explanation:

Step1: Calculate left - hand limit

We find $\lim_{x
ightarrow2^{-}}g(x)$. Since $x
ightarrow2^{-}$ (approaching 2 from the left, so $x\leq2$), $g(x)=(x - 2)^2$. Then $\lim_{x
ightarrow2^{-}}g(x)=\lim_{x
ightarrow2^{-}}(x - 2)^2=(2 - 2)^2 = 0$.

Step2: Calculate right - hand limit

We find $\lim_{x
ightarrow2^{+}}g(x)$. Since $x
ightarrow2^{+}$ (approaching 2 from the right, so $x>2$), $g(x)=2 - x^2$. Then $\lim_{x
ightarrow2^{+}}g(x)=\lim_{x
ightarrow2^{+}}(2 - x^2)=2-2^2=2 - 4=-2$.

Step3: Check continuity condition

For a function to be continuous at $x = a$, $\lim_{x
ightarrow a^{-}}g(x)=\lim_{x
ightarrow a^{+}}g(x)=g(a)$. Here, $\lim_{x
ightarrow2^{-}}g(x)=0$ and $\lim_{x
ightarrow2^{+}}g(x)=-2$. Since $\lim_{x
ightarrow2^{-}}g(x)
eq\lim_{x
ightarrow2^{+}}g(x)$, the function $g(x)$ is not continuous at $x = 2$.

Answer:

B. No