Is the function continuous $~x=1~$, and is it differentiable in $~x=1~$?

Question

$$f(x)= \begin{cases}(x-1)\left(1+ \sin\left(\dfrac{1}{x^2-1}\right)\right) & x\neq 1 \\ 0 & x=1 \end{cases}$$

Is the function continuous if I just show that the limit to the first expression is $~0~$ when $~x=1~$. I have to use the definition of the derivative to find out if it is differentiable in $~x=1~$, but I don't know how to do that with a combined function. I have tried to use the definition on the first function: $~(x-1)~$, but I get a big and messy expression.

I really appreciate some help.

Answer 1

HINT

One possible approach consists in making use of the squeeze theorem. Precisely, observe that \begin{align*} (x-1)\sin\left(\frac{1}{x^{2}-1}\right)\xrightarrow{\text{$x \rightarrow 1$}} 0 \end{align*}

That is because \begin{align*} \lim_{x\rightarrow 1} (x-1) = 0\quad\text{and}\quad\left|\sin\left(\frac{1}{x^{2}-1}\right)\right| \leq 1 \end{align*}

Can you proceed from here as to the derivative?

Answer 2

For the derivative, use the limit definition (observing that $f(1)=0$)

$$f'(1)=\lim_{x\to 1}\frac{f(x)-f(1)}{x-1}=\lim_{x\to 1}\frac{(x-1)(1+ \sin\left(\dfrac{1}{x^2-1}\right))-0}{x-1}=\lim_{x\to 1}\Bigg(1+ \sin\left(\dfrac{1}{x^2-1}\right)\Bigg)$$

and then observe that plotting $y=\Bigg(1+ \sin\left(\dfrac{1}{x^2-1}\right)\Bigg)$ forms

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To think about what happens as $x=1$ in the limit above, consider the more familiar problem

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

which has the following plot

As $x\to 0$, $\frac{1}{x}$ increases without bound. Thus in any open interval containing $0$ there will be values of $x$ such that $\frac{1}{x}$ is a multiple of $2 \pi$, values of $x$ such that $\frac{1}{x}$ is $\frac{\pi}{2}$ more than a multiple of $2 \pi$, and values of $x$ such that $\frac{1}{x}$ is $\frac{3 \pi}{2}$ more than a multiple of $2 \pi$. The sine of these values of $x$ will be $0, 1$ and $-1$ respectively.

Therefore, $\sin\left(\frac{1}{x}\right)$ oscillates "an infinite number of times" between $1$ and $-1$ in any neighborhood of $x \to 0$. Thus, the limit doesn't exist.

What does a similar analysis allow you to conclude about

$$\lim_{x\to 1}\Bigg(1+ \sin\left(\dfrac{1}{x^2-1}\right)\Bigg)?$$

Answer 3

Intuitively, at $x=1$ the function is the product of $x-1$, which tends to zero continuously, and an oscillating but bounded factor (in $[0,2]$). Hence the function is continuous (more rigorously, you can squeeze to $0$).

The analytical derivative is

$$((x-1)f(x))'=f(x)+(x-1)f'(x)$$ and is not defined because of $f(1)$.