Help with limits and a derivative

[Kolika28]

Homework Statement

$f(x) = \begin{cases} (x-1)(1+sin(\frac{1}{x^2-1})) & \quad \text{if } x \text{ is not equal to 1}\\ 0 & \quad \text{if } x \text{ is equal to 1} \end{cases}$

Is the function continuous and differentiable in x=1? Use the definition of the derivative.

Relevant Equations

The defintion of the derivative

I don't understand my textbook, so I simply don't understand how to solve this math problem. I really appreciate some help!

 

[fresh_42]

Are you sure the definition of the function isn't split at point $x=1$ instead of $x=0$?

 

[Kolika28]

Are you sure the definition of the function isn't split at point $x=1$ instead of $x=0$?

Of course, my mistake! I will fix it now!

 

[fresh_42]

I would check continuity first, since if it wasn't then it's also not differentiable. However, the problem asks you to use the limit of the derivative. So start at this point: what does differentiability at $x=1$ mean, i.e. which formalism of differentiability do you use?

 

[Kolika28]

So start at this point: what does differentiability at x=1x=1x=1 mean, i.e. which formalism of differentiability do you use?

By formalism do you mean definition? I use $\lim_{h\to 0}\frac{f(x+h)-f(x)}{h}$

 

[fresh_42]

Yes. You can write it as $\lim_{h \to 0} \dfrac{f(x_0+h)-f(x_0)}{h}$ or as $\lim_{x \to x_0} \dfrac{f(x)-f(x_0)}{x-x_0}$ or with Weierstrass as $f(x_0 + h) = f(x_0) + J(h) + r(h)$. In the first version we get
$$
\lim_{h \to 0} \dfrac{(x+h-1)(1+\sin(\frac{1}{(x+h)^2-1}))-(x-1)(1+\sin(\frac{1}{x^2-1}))}{h}
$$
So the difficulty is to manage the sine term. As it is bounded between $\pm 1$, taking the absolute value of the expression could work.

Do you have an idea which way it turns out?

 

[Kolika28]

Yes. You can write it as $\lim_{h \to 0} \dfrac{f(x_0+h)-f(x_0)}{h}$ or as $\lim_{x \to x_0} \dfrac{f(x)-f(x_0)}{x-x_0}$ or with Weierstrass as $f(x_0 + h) = f(x_0) + J(h) + r(h)$. In the first version we get
$$
\lim_{h \to 0} \dfrac{(x+h-1)(1+\sin(\frac{1}{(x+h)^2-1}))-(x-1)(1+\sin(\frac{1}{x^2-1}))}{h}
$$
So the difficulty is to manage the sine term. As it is bounded between $\pm 1$, taking the absolute value of the expression could work.

Do you have an idea which way it turns out?

I've tried the last thing you wrote me , but I struggle to get around the sin-function. I know that it is bounded but I don't know how to use it. And no, I don't have an idea which way it turns out. Thanks for helping my by the way, I really struggle with this. I'm a slow learner too, so it takes some time for me to fully understand.

 

[fresh_42]

My guess is that it is continuous at $x=1$ due to the decreasing amplitude, but not differentiable, so we have to show that the limit doesn't exist. The heuristic is
\begin{align*}
F(h) &:= \lim_{h \to 0} \dfrac{ (x+h-1) (1+\sin ( \frac{1}{(x+h)^2-1} ) ) - (x-1) (1+\sin ( \frac{1}{x^2-1} ) ) }{h} \\
&= \lim_{h \to 0} \left\{\left( \dfrac{x}{h}-\dfrac{1}{h}+1 \right)(1+sws)- \left( \dfrac{x}{h}-\dfrac{1}{h} \right)(1+sws')\right\} \\
&= \lim_{h \to 0} (1+sws'')
\end{align*}
where $sws$ means something that shivers, i.e. we have terms which oscillate between $\pm 1$, so the limit cannot be fixed. Now how do we make this rigorous? The limit definition is:

$F(h) \stackrel{h \to 0}{\longrightarrow} L$ iff for any $\varepsilon>0$ there is a $\delta > 0$ such that
$$
0< |h|< \delta \;\Longrightarrow\; |F(h)-L|< \varepsilon
$$
Unfortunately we cannot use $L=1$ which is what we would do in case of continuity, but here we have the entire quotient as function in $h$ and no idea what the limit would be. Hence we have to assume any fixed number $L$.

Can you negate the statement such that it becomes a definition of a non existing limit?

 

[fresh_42]

Do you think it would be ok to differentiate $f(x)$ at a point $x\neq 1$ and considering the derivative then, instead of from the start?

 

[Delta2]

I think its better to apply the definition $\lim_{x \to 1}\frac{f(x)-f(1)}{x-1}=\frac{(x-1)(1+\sin(\frac{1}{x^2-1}))-0}{x-1}$ since we are given that $f(1)=0$, and then all we are left with is the limit $$\lim_{x \to 1}\sin\frac{1}{x^2-1}$$ which doesn't exist ( I mean its a well known result that the limit of a periodic function does not exist at infinity).

 

[Kolika28]

Do you think it would be ok to differentiate $f(x)$ at a point $x\neq 1$ and considering the derivative then, instead of from the start?

I got it now, it took some time but I finally solved it! Thank you!