Consider the following in respect of the function $f(x) = \left\{\begin{matrix}2+ x, & x \geq 0\\ 2 - x, & x < 0\end{matrix}\right.$

1. $\displaystyle \lim_{x \rightarrow 1} f(x)$ does not exist.

2. f(x) is differentiable at x = 0.

3. f(x) is continuous at x = 0.

Which of the above statements is/are correct?

1. $\lim _{x \rightarrow 1} f(x)$ does not exist.

- For $x \geq 0, f(x)=2+x$. As $x$ approaches 1 from the right, $f(x)$ approaches $2+1=3$.

- Since $x=1$ is in the domain where $x \geq 0$, we only need to consider this part.

- Therefore, $\lim _{x \rightarrow 1} f(x)=3$, which exists.

Conclusion: This statement is incorrect.

2. $f(x)$ is differentiable at $x=0$.

- To check differentiability at $x=0$, we need to ensure that the left-hand derivative and the righthand derivative at $x=0$ are equal.

For $x \geq 0, f(x)=2+x$, so $f^{\prime}(x)=1$.

For $x<0, f(x)=2-x$, so $f^{\prime}(x)=-1$.

At $x=0$, the left-hand derivative is -1 and the right-hand derivative is 1 .

Since $-1 \neq 1, f(x)$ is not differentiable at $x=0$.

3. $f(x)$ is continuous at $x=0$.

To check continuity at $x=0$, we need to ensure that $\lim _{x \rightarrow 0} f(x)=f(0)$.

For $x \geq 0, f(x)=2+x$, so $f(0)=2$.

For $x<0, f(x)=2-x$, so as $x$ approaches 0 from the left, $f(x)$ approaches 2 .

Therefore, $\lim _{x \rightarrow 0} f(x)=2=f(0)$.

Final Answer:

Only statement 3 is correct.