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Evaluating undefined limits

  • Thread starter baki
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  • #1
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1. Evaluate



2. lim χ→0 [((χ+1)^1/3) -1] / χ



3. The hint is [a^2 + b^2 = (a+b)(a^2 + ab + b^2)].
But i don't get it how can that help me solve this because it is a cube root and not to the power of 3.


Without thinking about the hint, I have attempted it and i think it is undefined for the function.
Say f(x) is the function, so f(0) is undefined making the limit as x approaches 0 undefined as well.
is it correct??
 
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Answers and Replies

  • #2
Ray Vickson
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1. Evaluate



2. lim χ→0 [((χ+1)^1/3) -1] / χ



3. The hint is [a^2 + b^2 = (a+b)(a^2 + ab + b^2)].
But i don't get it how can that help me solve this because it is a cube root and not to the power of 3.


Without thinking about the hint, I have attempted it and i think it is undefined for the function.
Say f(x) is the function, so f(0) is undefined making the limit as x approaches 0 undefined as well.
is it correct??
The hint is incorrect: a^2 + b^2 has 'a' and 'b' to the second power, while the right-hand-side has them to the third power. A correct hint would be a^3 - b^3 = (a-b)(a^2 + ab + b^2).

Think of setting y = (x+1)^(1/3), so the numerator of your expression is y-1. Can you see how to relate y-1 to y^3 - 1? Can you see how to write the denominator (x) in terms of y? Can you see what happens to y when x --> 0?

RGV
 
  • #3
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Think of setting y = (x+1)^(1/3), so the numerator of your expression is y-1. Can you see how to relate y-1 to y^3 - 1? Can you see how to write the denominator (x) in terms of y? Can you see what happens to y when x --> 0?
That's one possibility and it works fine.
But I prefer to multiply numerator and denominator with [itex]a^2+ab+b^2[/itex] for suitable a and b. What do you get if you use the (correct version of the) hint?
 
  • #4
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sorry I was wrong, the hint is: a^3 - b^3 = (a-b)(a^2 + ab + b^2)

Ok, I did L'hopitals rule to derive the formula, until I am able to get a value from substituting zero in place of x. My answer came to 1/3. But still the hint had nothing to do with it.

Are you saying for me to do the opposite operation??
 
  • #5
SammyS
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sorry I was wrong, the hint is: a^3 - b^3 = (a-b)(a^2 + ab + b^2)

Ok, I did L'hopitals rule to derive the formula, until I am able to get a value from substituting zero in place of x. My answer came to 1/3. But still the hint had nothing to do with it.

Are you saying for me to do the opposite operation??
Here is how the hint might be used:

The numerator [itex]\sqrt[3]{x+1}-1[/itex]. If you could get an equivalent expression with a numerator of [itex]\displaystyle \left(\sqrt[3]{x+1}\right)^3-1^3\,,[/itex] that would allow major simplification including cancelling a factor that goes to zero in the numerator & denominator.

Use the hint to find out what you need to multiply [itex]\sqrt[3]{x+1}-1[/itex] by in order to get [itex]\displaystyle \left(\sqrt[3]{x+1}\right)^3-1^3\ .[/itex]

By the way, welcome to PF!
 
  • #6
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so then I would have to multiply it with ([itex]\sqrt[3]{x+1}[/itex])^2 + [itex]\sqrt[3]{x+1}[/itex] +1
and then do the same with the numerator??
is it??

and afterwards derive it???
 
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  • #7
SammyS
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so then I would have to multiply it with ([itex]\sqrt[3]{x+1}[/itex])^2 + [itex]\sqrt[3]{x+1}[/itex] +1
and then do the same with the numerator??
is it??

and afterwards derive it???
No, you should not need to use L'Hôpital's rule.

Just multiply the numerator & denominator by [itex]\displaystyle \left(\sqrt[3]{x+1\ }\right)^2+\sqrt[3]{x+1\ }+1\ .[/itex] Then simplify.
 
  • #8
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oh YES

solved!

other than simplifying we can just substitute 0, right?
 

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