(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.
$\displaystyle{{\lim _{x \to 1}}\dfrac{{{x^{(1/3)}} - 1}}{{{x^{(1/4)}} - 1}}\overbrace = ^H{\lim _{x \to 1}}\dfrac{{(1/3){x^{( - 2/3)}}}}{{(1/4){x^{( - 3/4)}}}}}$
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 $\displaystyle x = y^{12}.$
Then $\displaystyle x^{1/3} = y^4.$
And $\displaystyle x^{1/4} = y^3.$
As x approaches 1, y also approaches 1.
$\displaystyle \displaystyle\lim_{y\to 1} \dfrac{y^4 - 1}{y^3 - 1} = $
$\displaystyle \displaystyle\lim_{y\to 1} \dfrac{(y - 1)(y^3 + y^2 + y + 1)}{(y - 1)(y^2 + y + 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 $\displaystyle y$ to infer the existence (and value) of the limit in $\displaystyle x$. Thus the relevant implication is that as $\displaystyle y \to 1$, $\displaystyle x \to 1^+$. I guess we ought to do a little more work to prove the left-sided limit.