Let f(x) = begin{cases} 0, & x=0 2-x, & 0

Question

(D) For $x=0$: $\lim_{x ightarrow 0^{+}} f(x) = \lim_{x ightarrow 0^{+}} (2-x) = 2$. Since $f(0) = 0$,$\lim_{x ightarrow 0^{+}} f(x) 
eq f(0)$,

Vedclass · Accepted Answer

$f$ is continuous at $x=2$

Vedclass · Answer

(D) For $x=0$: $\lim_{x ightarrow 0^{+}} f(x) = \lim_{x ightarrow 0^{+}} (2-x) = 2$. Since $f(0) = 0$,$\lim_{x ightarrow 0^{+}} f(x) 
eq f(0)$,so $f$ is not right continuous at $x=0$. For $x=1$: $\lim_{x ightarrow 1^{-}} f(x) = \lim_{x ightarrow 1^{-}} (2-x) = 1$. Since $f(1) = 2$,$\lim_{x ightarrow 1^{-}} f(x) 
eq f(1)$,so $f$ is not left continuous at $x=1$. $\lim_{x ightarrow 1^{+}} f(x) = \lim_{x ightarrow 1^{+}} (\frac{1}{2}-x) = -\frac{1}{2}$. Since $f(1) = 2$,$\lim_{x ightarrow 1^{+}} f(x) 
eq f(1)$,so $f$ is not right continuous at $x=1$. For $x=2$: $\lim_{x ightarrow 2^{-}} f(x) = \lim_{x ightarrow 2^{-}} (\frac{1}{2}-x) = \frac{1}{2}-2 = -\frac{3}{2}$. $\lim_{x ightarrow 2^{+}} f(x) = \lim_{x ightarrow 2^{+}} (-\frac{3}{2}) = -\frac{3}{2}$. $f(2) = -\frac{3}{2}$. Since $\lim_{x ightarrow 2^{-}} f(x) = \lim_{x ightarrow 2^{+}} f(x) = f(2) = -\frac{3}{2}$,$f$ is continuous at $x=2$.

Let $f(x) = \begin{cases} 0, & x=0 \\ 2-x, & 0 < x < 1 \\ 2, & x=1 \\ \frac{1}{2}-x, & 1 < x < 2 \\ \frac{-3}{2}, & x \geq 2 \end{cases}$ then which of the following is true?

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