Why does not match?

Rubén

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Dec 8, 2018
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Anyone knows why this can happen on c)?
And what is the result of b), I don't get it solved.
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Thanks.
 
Anyone knows why this can happen on c)?
And what is the result of b), I don't get it solved.
View attachment 10894
Thanks.
What have you tried for part b? Where did you get stuck. Please show us your work so we can where you need help.
I would take the limit of (f(a+h)-f(a))/h where a=0.
 
What expression did you manage for part a?

Why would they be the same? What sort of singularity have we at x = 0?
The first, x <> 0.
Avoidable discontinuity.
I'm spanish, sorry if I don't speak correctly.
 
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What have you tried for part b? Where did you get stuck. Please show us your work so we can where you need help.
I would take the limit of (f(a+h)-f(a))/h where a=0.
I did, but I'm not able to remove the h from the denominator.
How can I write maths here?
 
I did, but I'm not able to remove the h from the denominator.
How can I write maths here?
The easiest way for you to post your work is by attaching a picture of it. You try to type it out but it can get messy if you do not now TEX.
 
I did it!
The limit of the incremental quotient gives me 1.
Why is it different of substituting in the normal derivative the x by 0?

I'm just super unhappy with this question.

Why would I substitute x = 0 into the expression for the derivative that I know is not defined for x = 0?!

Thus:

Limit ==> Value
Substitution ==> Not Defined, so I didn't do it.

Not much to compare.
 
The function, for x not 0, is \(\displaystyle f(x)= x+ x^2 sin(1/x)\) (this does not exit for x= 0 but the limit, as x goes to 0, is 0 so the given function is continuous at x= 0). It's derivative is \(\displaystyle f'(x)= 1+ 2x sin(1/x)+ x^2 cos(1/x)(-1/x^2)= 1+ 2x sin(1/x)- cos(1/x)\). That does not exist for x= 0 nor does it have a limit as x goes to 0.

The derivative of f at x= 0 is given by \(\displaystyle \lim_{h\to 0}\frac{f(0+h)- f(0)}{h}= \lim_{h\to 0}\frac{h+ h^2 sin(1/h)- 0}{h}= \lim_{h\to 0} 1+ h sin(1/h)= 1\).

(c) The function, f, is differentiable at x= 0 but the derivative is not continuous there.

(One can show that, while the derivative of a function is not necessarily continuous on a given interval, it must satisfy the "intermediate value property": "if f'(a)= P and f'(b)= Q, then f' must take on every value between P and Q for some x between a and b".
 
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