A function f(x)f(x)f(x) is defined as f(x)={2x+1x<15x=1x2+2x>1f(x) = \begin{cases} 2x + 1 & x < 1 \\ 5 & x = 1 \\ x^2 + 2 & x > 1 \end{cases}f(x)=⎩⎨⎧2x+15x2+2x<1x=1x>1. Is f(x)f(x)f(x) continuous at x=1x = 1x=1?
Yes, because limx→1f(x)=f(1)\lim_{x \to 1} f(x) = f(1)limx→1f(x)=f(1)
No, because limx→1−f(x)≠limx→1+f(x)\lim_{x \to 1^-} f(x) \neq \lim_{x \to 1^+} f(x)limx→1−f(x)=limx→1+f(x)
No, because limx→1f(x)≠f(1)\lim_{x \to 1} f(x) \neq f(1)limx→1f(x)=f(1)
Yes, because the function is defined at x=1x=1x=1