# Math Help - Algebra after using l'Hopital's Rule

1. ## Algebra after using l'Hopital's Rule

I would like to ask another question.
I had to calculate a limit using l'Hospital's rule.
I know that I got the correct numerical answer but unfortunately my answer has a different form from that of the professor.

Professor's answer: $\frac{3^\frac{5}{3}}{2^\frac{3}{2}}$

$\frac{\frac{-1}{2}*2^\frac{-1}{2}}{\frac{-1}{3}*3^\frac{-2}{3}}$

How can I change my form in the better one which he uses?

2. Originally Posted by Sebastian de Vries
I would like to ask another question.
I had to calculate a limit using l'Hospital's rule.
I know that I got the correct numerical answer but unfortunately my answer has a different form from that of the professor.

Professor's answer: $\frac{3^\frac{5}{3}}{2^\frac{3}{2}}$

$\frac{\frac{-1}{2}*2^\frac{-1}{2}}{\frac{-1}{3}*3^\frac{-2}{3}}$

How can I change my form in the better one which he uses?
Multiply top and bottom by -1 to get rid of the minus signs, so now you have:

$\frac{\frac{1}{2}\times 2^\frac{-1}{2}}{\frac{1}{3}\times 3^\frac{-2}{3}}$

Then observe that $1/2=2^{-1}$ and $1/3=3^{-1}$ to get:

$\frac{2^{-1}\times 2^\frac{-1}{2}}{3^{-1}\times 3^\frac{-2}{3}}$

Now laws of indices give:

$\frac{ 2^\frac{-1-1}{2}}{ 3^\frac{-1-2}{3}} =\frac{ 2^\frac{-3}{2}}{ 3^\frac{-5}{3}}$

Now flip the whole thing upside down to get rid of the negative powers to get the Prof's answer.

.

3. $\lim_{ x\to1}\frac{(x+1)^\frac{1}{2}-(2x)^\frac{1}{2}}{(x+2)^\frac{1}{3}-(2x+1)^\frac{1}{3}}$

I attempted to type the asked limit but as you can see the software can't execute the code (my mistake of course).

Mr F edit: You would see that I had fixed the latex if you stopped re-editing and hence undoing my editing!
Sebastian: sorry.

4. Originally Posted by Sebastian de Vries
$\lim_{ x\to1}\frac{(x+1)^\frac{1}{2}-(2x)^\frac{1}{2}}{(x+2)^\frac{1}{3}-(2x+1)^\frac{1}{3}}$

I attempted to type the asked limit but as you can see the software can't execute the code (my mistake of course).

If x approaches 1 your quotient will become $\dfrac00$. Therefore you should use de l'Hopitals rule:
$\lim_{ x\to1}\dfrac{(x+1)^\frac{1}{2}-(2x)^\frac{1}{2}}{(x+2)^\frac{1}{3}-(2x+1)^\frac{1}{3}} = \lim_{x\to1}\dfrac{\dfrac{\sqrt{x}-\sqrt{2} \sqrt{x+1}}{2\sqrt{x}\sqrt{x+1}}}{\dfrac{(2x+1)^{\ frac23} - 2(x+2)^{\frac23}}{3(x+2)^{\frac23}(2x+1)^{\frac23} }}$
$\lim_{ x\to1}\dfrac{(x+1)^\frac{1}{2}-(2x)^\frac{1}{2}}{(x+2)^\frac{1}{3}-(2x+1)^\frac{1}{3}} = \dfrac34 \cdot \sqrt{2} \cdot \sqrt[3]{9}\approx 2.206$