Help with limits and a derivative

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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!
 

Answers and Replies

  • #2
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Are you sure the definition of the function isn't split at point ##x=1## instead of ##x=0##?
 
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  • #3
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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!
 
  • #4
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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?
 
  • #5
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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} ##
 
  • #6
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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?
 
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  • #7
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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.
 
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  • #8
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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?
 
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  • #9
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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?
 
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  • #10
Delta2
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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).
 
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  • #11
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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!
 
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