Let $f(x)=\begin{cases}\dfrac{|x|}{x},&x\ne0\\1,&x=0\end{cases}$ and $g(x)=\begin{cases}x\sin\left(\dfrac{1}{x}\right),&x\ne0\\0,&x=0\end{cases}$

Then at the origin, which one of the following is true?

We are given:

$f(x)=\begin{cases}\dfrac{|x|}{x},&x\ne0\\[6pt]1,&x=0\end{cases}$

and

$g(x)=\begin{cases}x\sin\left(\dfrac{1}{x}\right),&x\ne0\\[6pt]0,&x=0\end{cases}$

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Let's first simplify $f(x)$ for $x \ne 0$:

• When $x > 0$: $f(x) = \dfrac{x}{x} = 1$
• When $x < 0$: $f(x) = \dfrac{-x}{x} = -1$
• At $x = 0$: $f(0) = 1$

So $f(x)$ outputs $+1$ for all $x \geq 0$ and $-1$ for $x < 0$.

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Checking continuity of $f(x)$ at $x = 0$:

$\displaystyle\lim_{x \to 0^+} f(x) = 1$

$\displaystyle\lim_{x \to 0^-} f(x) = -1$

Since the left-hand limit $\neq$ right-hand limit, $\displaystyle\lim_{x \to 0} f(x)$ does not exist.

Therefore, $f(x)$ is not continuous at $x = 0$.

Since $f(x)$ is not continuous at $x = 0$, it cannot be differentiable there either (continuity is a necessary condition for differentiability).

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Checking continuity of $g(x)$ at $x = 0$:

We need $\displaystyle\lim_{x \to 0} g(x) = g(0) = 0$.

We know that $-1 \leq \sin\left(\dfrac{1}{x}\right) \leq 1$ for all $x \ne 0$.

Multiplying through by $|x|$:

$-|x| \leq x\sin\left(\dfrac{1}{x}\right) \leq |x|$

As $x \to 0$, both $-|x| \to 0$ and $|x| \to 0$.

By the Squeeze (Sandwich) Theorem:

$\displaystyle\lim_{x \to 0} x\sin\left(\dfrac{1}{x}\right) = 0 = g(0)$

Therefore, $g(x)$ is continuous at $x = 0$.

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Checking differentiability of $g(x)$ at $x = 0$:

Using the definition of the derivative:

$g'(0) = \displaystyle\lim_{h \to 0} \dfrac{g(h) - g(0)}{h}$

$= \displaystyle\lim_{h \to 0} \dfrac{h\sin\left(\dfrac{1}{h}\right) - 0}{h}$

$= \displaystyle\lim_{h \to 0} \sin\left(\dfrac{1}{h}\right)$

As $h \to 0$, $\sin\left(\dfrac{1}{h}\right)$ oscillates infinitely between $-1$ and $+1$. This limit does not exist.

Therefore, $g(x)$ is not differentiable at $x = 0$.

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Summary:

	Continuous at $x = 0$?	Differentiable at $x = 0$?
$f(x)$	No	No
$g(x)$	Yes	No

The correct answer is: $f$ is not continuous but $g$ is continuous and not differentiable at the origin.