Question

Which of the following statements are true for the function $f\left(x\right)$ defined for $-1\le x\le 3$ in the figure shown?

• A. $\underset{x\to -1^{+}}{lim}f\left(X\right)=1$
• B. $\underset{x\to 2}{lim}f\left(X\right)$ does not exist
• C. $\underset{x\to 1^{-}}{lim}f\left(X\right)=2$
• D. $\underset{x\to 0^{+}}{lim}f\left(X\right)=\underset{x\to 0^{-}}{lim}f\left(X\right)$

Answer 1

Answer: (A), (B), (C), (D)

The correct options are
A $\underset{x\to -1^{+}}{lim}f\left(X\right)=1$
B $\underset{x\to 2}{lim}f\left(X\right)$ does not exist
C $\underset{x\to 1^{-}}{lim}f\left(X\right)=2$
D $\underset{x\to 0^{+}}{lim}f\left(X\right)=\underset{x\to 0^{-}}{lim}f\left(X\right)$

From the picture it is clear that when $x\to 1^{+}$, $f\left(x\right)=1$ i.e., we are getting a neighbour-hood $1<x\le 3$ where $f\left(x\right)=1$.

So, $\underset{x\to 1^{+}}{lim}f\left(x\right)=1$.

So, option (A) is correct.

Also as $x\to 1^{-}$, $f\left(x\right)=2$, so, option (C) is also correct.

(C) is also corrcet because from the picture it is clear that we are getting a neighbou-hood of $2$ where $f\left(x\right)=1$ and $f\left(x\right)=2$ as well, so $\underset{x\to 2}{lim}f\left(x\right)$ does not exist.

Again (D) is also correct as we have a small neighbour-hood of $0$ where the limits of $f\left(x\right)$ exists.