Question

question 2 (1 point) saved let $f(x)=\begin{cases}\frac{x^{2}-4}{x - 2},&x
eq2\\1,&x = 2end{cases}$ which of the following statements about $f(x)$ are true? $f$ has a limit at $x = 2$. $f(2)=1$ $f$ is not differentiable at $x = 2$. $f$ has a removable continuity. $f$ is continuous at $x = 2$.

Explanation:

Step1: Simplify the non - constant part of the function

For $x
eq2$, $f(x)=\frac{x^{2}-4}{x - 2}=\frac{(x + 2)(x - 2)}{x - 2}=x + 2$.

Step2: Calculate the limit as $x$ approaches 2

$\lim_{x
ightarrow2}f(x)=\lim_{x
ightarrow2}(x + 2)=4$. Since $f(2)=1$, the function has a removable discontinuity at $x = 2$.

Step3: Analyze differentiability

A function is differentiable at a point if it is continuous at that point. Since $f(x)$ is not continuous at $x = 2$ (because $\lim_{x
ightarrow2}f(x)=4
eq f(2)=1$), it is not differentiable at $x = 2$.

Answer:

$f$ has a limit at $x = 2$. True, $\lim_{x
ightarrow2}f(x)=4$.
$f(2)=1$. True, as defined in the piece - wise function.
$f$ is not differentiable at $x = 2$. True, because it is not continuous at $x = 2$.
$f$ has a removable continuity. True, since $\lim_{x
ightarrow2}f(x)$ exists but $\lim_{x
ightarrow2}f(x)
eq f(2)$.
$f$ is continuous at $x = 2$. False, since $\lim_{x
ightarrow2}f(x)=4$ and $f(2)=1$.