1. ## Find the limit

(x^(1/3)-1)/(x^(1/4)-1) find the limit as x approaches 1

2. ## Re: Find the limit

$(x^{\frac13}-1)(x^{\frac23}+x^{\frac13}+1)=x-1$ so you can multiply numerator and denominator by the second factor to simplify the numerator.

You can do a similar trick with the denominator.

3. ## Re: Find the limit

Originally Posted by GodSalamence
(x^(1/3)-1)/(x^(1/4)-1) find the limit as x approaches 1
I have never been a fan of telling a student to use L'Hopital's Rule. But in this case it seems justified, especially if one is algebraically challenged.

$\displaystyle{{\lim _{x \to 1}}\dfrac{{{x^{(1/3)}} - 1}}{{{x^{(1/4)}} - 1}}\overbrace = ^H{\lim _{x \to 1}}\dfrac{{(1/3){x^{( - 2/3)}}}}{{(1/4){x^{( - 3/4)}}}}}$

4. ## Re: Find the limit

Originally Posted by Archie
$(x^{\frac13}-1)(x^{\frac23}+x^{\frac13}+1)=x-1$ so you can multiply numerator and denominator by the second factor to simplify the numerator.

You can do a similar trick with the denominator.
I know you make it a practice to avoid using L'Hopital's Rule in doing these limit problems, but what you recommend
here is relatively awkward, especially because of the fractional exponents. Let's still avoid L'Hopital's Rule.

12 is the LCD of the fractions used for the exponents.

Let $x = y^{12}.$

Then $x^{1/3} = y^4.$

And $x^{1/4} = y^3.$

As x approaches 1, y also approaches 1.

$\displaystyle\lim_{y\to 1} \dfrac{y^4 - 1}{y^3 - 1} =$

$\displaystyle\lim_{y\to 1} \dfrac{(y - 1)(y^3 + y^2 + y + 1)}{(y - 1)(y^2 + y + 1)} =$

5. ## Re: Find the limit

Originally Posted by greg1313
As x approaches 1, y also approaches 1.
It's a minor point, but one that is usually missed and is occasionally important: the implication here is in the wrong direction. You are using the existence (and value) of the limit in $y$ to infer the existence (and value) of the limit in $x$. Thus the relevant implication is that as $y \to 1$, $x \to 1^+$. I guess we ought to do a little more work to prove the left-sided limit.

6. ## Re: Find the limit

Originally Posted by Archie
Thus the relevant implication is that as $y \to 1$, $x \to 1^+$.
This bit is wrong. Of course $y^{12}$ can be smaller than $1$. But the general point still stands.

7. ## Re: Find the limit

If $y = x^{1/12}$, then as x approaches 1, so that will mean that y
will approach 1, as they are equal.

8. ## Re: Find the limit

As I said, it's a minor point in this case. But if the limit were zero, we'd have a problem because the equality $x=y^{12}$ would hold only on one side of the limit point.

9. ## Re: Find the limit

Originally Posted by Archie
As I said, it's a minor point in this case. But if the limit were zero, we'd have a problem because the equality $x=y^{12}$ would hold only on one side of the limit point.
But the limit is not "as x approaches zero," so why bring that up? That doesn't have a bearing on what I'm doing.
I already knew/know as x approaches 1 from the left and from the right, that $x^{1/12}$ approaches 1.