Let $R$ denote the set of all real numbers. Let

$f:R\to R$

be an arbitrary function and let

$g:R\to R$

be the function defined by

$g\left(x\right)=xf\left(x\right),\forall x\in R$

Then which of the following statements is (are) TRUE?

Correct Options:
1. If $f$ is continuous at $x=0$, then $g$ is differentiable at $x=0$
2. If $g$ is differentiable at $x=0$, then $\underset{x\to 0}{lim}f(x)$ exists

Step-by-Step Explanation:

We are given that $g\left(x\right)=xf\left(x\right)$ for all $x\in R$.

First, note that evaluating $g\left(0\right)$ gives:

$g\left(0\right)=0·f\left(0\right)=0$

Now let us analyze differentiability of $g\left(x\right)$ at $x=0$ using the first definition of the derivative:

$g^{\prime }\left(0\right)=\underset{h\to 0}{lim}\frac{g(0+h)-g\left(0\right)}{h}$

Substitute $g\left(h\right)=hf\left(h\right)$ and $g\left(0\right)=0$ into the limit:

$g^{\prime }\left(0\right)=\underset{h\to 0}{lim}\frac{hf\left(h\right)-0}{h}=\underset{h\to 0}{lim}f\left(h\right)$

Thus, $g\left(x\right)$ is differentiable at $x=0$ if and only if $\underset{x\to 0}{lim}f(x)$ exists as a finite real number.

Analysis of the Statements:

1. Statement 2: "If $f$ is continuous at $x=0$, then $g$ is differentiable at $x=0$"
If $f$ is continuous at $x=0$, then by definition of continuity, $\underset{x\to 0}{lim}f(x)=f(0)$ exists and is finite.
Since $\underset{x\to 0}{lim}f(x)$ exists, it follows directly from our derivation that $g^{\prime }\left(0\right)=f(0)$, so $g$ is differentiable at $x=0$.
Hence, this statement is TRUE.

2. Statement 4: "If $g$ is differentiable at $x=0$, then $\underset{x\to 0}{lim}f(x)$ exists"
If $g$ is differentiable at $x=0$, the limit definition requires $g^{\prime }\left(0\right)=\underset{h\to 0}{lim}f\left(h\right)$ to exist as a finite number.
Therefore, $\underset{x\to 0}{lim}f(x)$ must exist.
Hence, this statement is TRUE.