Yes, substituting 0 would also show that the limit is undefined.

[baki]

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??

 

[Ray Vickson]

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

 

[micromass]

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 a^2+ab+b^2 for suitable a and b. What do you get if you use the (correct version of the) hint?

 

[baki]

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??

 

[SammyS]

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 \sqrt[3]{x+1}-1. If you could get an equivalent expression with a numerator of \displaystyle \left(\sqrt[3]{x+1}\right)^3-1^3\,, 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 \sqrt[3]{x+1}-1 by in order to get \displaystyle \left(\sqrt[3]{x+1}\right)^3-1^3\ .

By the way, welcome to PF!

 

[baki]

so then I would have to multiply it with (\sqrt[3]{x+1})^2 + \sqrt[3]{x+1} +1
and then do the same with the numerator??
is it??

and afterwards derive it?

 

[SammyS]

so then I would have to multiply it with (\sqrt[3]{x+1})^2 + \sqrt[3]{x+1} +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 \displaystyle \left(\sqrt[3]{x+1\ }\right)^2+\sqrt[3]{x+1\ }+1\ . Then simplify.

 

[baki]

oh YES

solved!

other than simplifying we can just substitute 0, right?