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How to evaluate the limit?

  • Thread starter MathewsMD
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  • #1
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Evaluate: f(x+h) - f(x) / h if f(x) = (x^2)sin(1/x)

I've tried evaluating, but only end up with 2xsin(1/x), and that is incorrect. I have also not learned L'Hospital's rule yet so any help would be great!
 

Answers and Replies

  • #2
Office_Shredder
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Is the question asking you to evaluate that expression for any function h, or is it asking you to calculate the limit as h goes to zero (i.e. find the derivative)?

Assuming the latter, you need to use the product rule
 
  • #3
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Is the question asking you to evaluate that expression for any function h, or is it asking you to calculate the limit as h goes to zero (i.e. find the derivative)?

Assuming the latter, you need to use the product rule

It is the latter. And you can only evaluate using limits.
 
  • #4
haruspex
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Pls post your attempt.
 
  • #5
Office_Shredder
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I think it's bizarre they want you to calculate this using the limit definition, but it is what it is. Like haruspex said, to help you we will need to see what calculations you did when you tried calculating the limit (we should be able to easily identify the point where you made your mistake)
 
  • #6
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I did:

=lim [[(x+h)^2]sin(1/x+h) - (x^2)sin(1/x)]/h
h→0

=lim [(x^2)sin(1/x+h) + 2xhsin(1/x+h) + (h^2)sin(1/x+h) - (x^2)sin(1/x)]/h
h→0

=[lim[(x^2)sin(1/x+h) - (x^2)sin(1/x)] + lim[2xhsin(1/x+h) + (h^2)sin(1/x+h)]]/lim h
h→0 h→0 h→0

= 0 + lim [[h (2xsin(1/x+h) + hsin(1/x+h)]/h]
h→0

= lim (2xsin(1/x+h) + hsin(1/x+h)]
h→0

= 2xsin(1/x)

If there are any errors in my steps, please point them out. This is not the correct answer (I believe), and if someone could show me the correct way to evaluate this limit, both when x = 0 and when x ≠ 0, that would be great. Thanks!

EDIT: The h→0 should be everywhere you see the lim notation. Sorry, I can't really format well here.
 
  • #7
Office_Shredder
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You seem to have made the claim that
[tex] \lim_{h\to \0} \frac{ x^2 \sin(1/(x+h)) - x^2 \sin(1/x)}{h} = 0 [/tex]

which is not correct.
 
  • #8
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I meant:


[tex] \lim_{h\to\ 0} {x^2 \sin(1/(x+h)) - x^2 \sin(1/x)} = 0 [/tex]


Then I assigned the limit to the denominator (h) and since the numerator was just addition, I did the same for the last two terms.
 
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  • #9
Office_Shredder
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I see what you are doing now - it was hard to read in your last post but you cannot move around limits and fractions like that. Once you get to this point

[tex] \frac{ x^2 \sin(1/(x+h)) - x^2 \sin(1/x) + 2xh\sin(1/(x+h)) + h^2 \sin(1/(x+h))}{h} [/tex]

You cannot re-write this as
[tex] x^2 \sin(1/(x+h)) - x^2 \sin(1/x) + \frac{2xh \sin(1/(x+h)) + h^2 \sin(1/(x+h)}{h} [/tex]
 
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  • #10
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[tex][ \lim_{h\to\ 0}{(x^2 \sin(1/(x+h)) - x^2 \sin(1/x)) + \lim_{h\to\ 0}(2xh\sin(1/(x+h)) + h^2 \sin(1/(x+h))}] ÷ \lim_{h\to\ 0}{h} [/tex]
 
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  • #11
Office_Shredder
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First, you still split the fraction in a way that you are not allowed to (by not having an h in the denominator of your first term). Second, you can only split the limit between the numerator and denominator like that if both limits exist and the denominator does not go to zero, which is clearly does. Your second fraction is 0/0, not 2xsin(1/x).
 
  • #12
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First, you still split the fraction in a way that you are not allowed to (by not having an h in the denominator of your first term). Second, you can only split the limit between the numerator and denominator like that if both limits exist and the denominator does not go to zero, which is clearly does. Your second fraction is 0/0, not 2xsin(1/x).
Okay. So what method would be best in this case to overcome this problem?

I don't think factoring is really an option here and I don't see a good term to use to rationalize.
 
  • #13
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Is there any path I should take or something I am missing that would help me further simplify this limit?
 
  • #14
Office_Shredder
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Try writing it as
[tex]\lim_{h\to 0} \frac{ x^2\sin(1/(x+h)) - x^2 \sin(1/x)}{h} + \lim_{h\to 0}\frac{ 2xh \sin(1/x) + h^2 \sin(1/x)}{h} [/tex]
 
  • #15
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Try writing it as
[tex]\lim_{h\to 0} \frac{ x^2\sin(1/(x+h)) - x^2 \sin(1/x)}{h} + \lim_{h\to 0}\frac{ 2xh \sin(1/x) + h^2 \sin(1/x)}{h} [/tex]

The right side would simplify to 2xsin(1/x) but it's the left side that I'm having trouble with. You can factor out the x^2 and then have the definition of the derivative of sin(1/x) which equals cos(1/x).
I know there are proofs to go from the left to the right side, but I was trying to understand and devise my own method I guess, which was the harder part.

Thanks for the help.
 
  • #16
Office_Shredder
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You can factor out the x^2 and then have the definition of the derivative of sin(1/x) which equals cos(1/x).
This is not quite right.

I know there are proofs to go from the left to the right side, but I was trying to understand and devise my own method I guess, which was the harder part.
I don't understand what you mean by going from the left to the right side... the left to right side of what I wrote? Those two halfs of the expression are fairly independent.
 
  • #17
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This is not quite right.



I don't understand what you mean by going from the left to the right side... the left to right side of what I wrote? Those two halfs of the expression are fairly independent.
Oh sorry. I initially wrote [tex]\lim_{h\to 0} \frac{\sin(1/(x+h)) - \sin(1/x)}{h} [/tex]

Took it out since I'm still learning to format and it wasn't coming out properly. Sorry for the poor formatting here.

And can't you factor the x^2 out of this limit?

[tex]\lim_{h\to 0} \frac{ x^2\sin(1/(x+h)) - x^2 \sin(1/x)}{h} [/tex]
 
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