Help with limits and a derivative

• Kolika28
In summary, the student struggled with understanding the definition of the function, and then struggled to differentiate the function at a point where it was not differentiable. He was eventually able to figure it out, thanks to a helpful person.
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!

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

berkeman
fresh_42 said:
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!

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?

fresh_42 said:
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} ##

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?

Last edited:
Kolika28
fresh_42 said:
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.

Delta2
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?

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?

Kolika28
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).

WWGD and Kolika28
fresh_42 said:
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!

WWGD and fresh_42

What is a limit?

A limit is the value that a function approaches as the input approaches a certain value. It is denoted by the notation lim f(x) or lim x->a.

How do you find the limit of a function?

To find the limit of a function, you can evaluate the function at values close to the point in question, and see what value the function approaches. Alternatively, you can use algebraic techniques such as factoring or rationalizing the denominator.

What is a derivative?

A derivative is a measure of how a function changes as its input changes. It is defined as the slope of the tangent line to the function at a specific point. It is denoted by the notation f'(x) or df/dx.

How do you find the derivative of a function?

To find the derivative of a function, you can use the limit definition of a derivative, which involves taking the limit of the difference quotient as the change in the input approaches zero. Alternatively, you can use differentiation rules, such as the power rule or the product and quotient rules.

Why are limits and derivatives important in science?

Limits and derivatives are important in science because they allow us to model and understand real-world phenomena, such as the rate of change of a chemical reaction or the velocity of an object. They are also fundamental concepts in calculus, which is a crucial tool for solving complex scientific problems.

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