A function is built from two different rules, one on each side of x=2x = 2x=2:
f(x)={x2−4x−2if x<25−xif x>2f(x) = \begin{cases} \dfrac{x^2 - 4}{x - 2} & \text{if } x < 2 \\ 5 - x & \text{if } x > 2 \end{cases}f(x)=⎩⎨⎧x−2x2−45−xif x<2if x>2
The function is not defined at x=2x = 2x=2 itself.
Which statement about fff near x=2x = 2x=2 is correct?
Select all that apply.
limx→2−f(x)=0\lim_{x \to 2^-} f(x) = 0limx→2−f(x)=0 and limx→2+f(x)=3\lim_{x \to 2^+} f(x) = 3limx→2+f(x)=3, so the two-sided limit does not exist.
limx→2−f(x)=4\lim_{x \to 2^-} f(x) = 4limx→2−f(x)=4 and limx→2+f(x)=3\lim_{x \to 2^+} f(x) = 3limx→2+f(x)=3, so the two-sided limit does not exist.
limx→2f(x)=4\lim_{x \to 2} f(x) = 4limx→2f(x)=4, since the left branch simplifies to x+2x + 2x+2.
limx→2−f(x)=3\lim_{x \to 2^-} f(x) = 3limx→2−f(x)=3 and limx→2+f(x)=4\lim_{x \to 2^+} f(x) = 4limx→2+f(x)=4, so the two-sided limit does not exist.