Proving functions as continuous

Question

I was tasked with proving these two functions are continuous:

$$g(x)=\begin{cases}3+\log(x), & x<e\\5, & x\ge e\end{cases}$$

$$j(x)=\begin{cases}x\sin\left(\frac{1}{x}\right), & x\ne 0\\0, & x=0\end{cases}$$

I know that a function is continuous if the limit of the function approaching from the left side is the same as the limit of the function approaching from the right side. I did some basic exercises and managed to do them quite easily. However, I am stumped at this particular exercise.

My take on this exercise: For $g,$ the limit approaching from the side is the $3+\log(e)$, which equals to $3.4342945$‬. The left side is simply $5$? Which means the function is not continuous.

As for $j,$ I have no idea what to do. Do I take the $0$ as the $x$? Then just plug $0$ into the $x \sin (1/x)$ twice and have them equal to each other, thus making the function continuous?

I shouldn't use graphs for this.

Answer 1

(a) It's the natural logarithm that's being used here. So, $\lim_{x\to e^-}g(x)=4\neq5=\lim_{x\to e^+}g(x)$. So, $g$ is not continuous at $e$.

(b) Since $(\forall x\in\mathbb R):-\lvert x\rvert\leqslant j(x)\leqslant\lvert x\rvert$, it follows from the squeeze theorem that $\lim_{x\to0}j(x)=0=j(0)$. Therefore, $j$ is continuous at $0$.

Answer 2

For part b)

$$\lim_{x \to 0} x \sin(\frac{1}{x})$$ $$=\lim_{x\to 0}\frac{ \sin(\frac{1}{x})}{\frac{1}{x}}$$ $$=0$$

Because $\sin (t) \in [-1,1] \quad \forall t$, and the denominator in the expression approaches $\infty$ as $x\to 0$

Therefore the function is continuous.