How to solve this limit

[juanpnav]

It can't be solve with L'hopital and I have no clue on how to do it.

 

[limiTS]

Start by putting them over a common denominator: $\displaystyle\frac{3}{1-x^{1/2}}-\frac{2}{1-x^{1/3}}=\frac{3\left(1-x^{1/3}\right)-2\left(1-x^{1/2}\right)}{\left(1-x^{1/2}\right)\left(1-x^{1/3}\right)}$

Then let $\displaystyle x=\lim_{h\to0}\left(1+h\right)$, and substitute in:

$\displaystyle\frac{3\left(1-\left(1+h\right)^{1/3}\right)-2\left(1-\left(1+h\right)^{1/2}\right)}{\left(1-\left(1+h\right)^{1/2}\right)\left(1-\left(1+h\right)^{1/3}\right)}$

Express the square root and the cube root as their binomial expansions:

$\displaystyle1-\left(1+h\right)^{1/2}=1-\left(1+\frac{h}{2}-\frac{h^2}{8}\right)=-\frac{h}{2}+\frac{h^2}{8}$
$\displaystyle1-\left(1+h\right)^{1/3}=1-\left(1+\frac{h}{3}-\frac{h^2}{9}\right)=-\frac{h}{3}+\frac{h^2}{9}$

Plug these expansions into the above.

After you do the algebra, you'll have $h^2$ as the dominant terms in both numerator and denominator. Cancel all the $h^2$ and then the answer is $\displaystyle\frac{1}{2}$

 

[MarkFL]

Do you mean you do not know how to apply L'Hopital or you have been instructed not to use it?

 

[juanpnav]

Do you mean you do not know how to apply L'Hopital or you have been instructed not to use it?

I mean I'm trying to help a student that it's just starting with limits and applying L'hopital would be seen as cheating. 10years I haven't done a limit so, yeah, need help with this one.

 

[juanpnav]

Start by putting them over a common denominator: $\displaystyle\frac{3}{1-x^{1/2}}-\frac{2}{1-x^{1/3}}=\frac{3\left(1-x^{1/3}\right)-2\left(1-x^{1/2}\right)}{\left(1-x^{1/2}\right)\left(1-x^{1/3}\right)}$

Then let $\displaystyle x=\lim_{h\to0}\left(1+h\right)$, and substitute in:

$\displaystyle\frac{3\left(1-\left(1+h\right)^{1/3}\right)-2\left(1-\left(1+h\right)^{1/2}\right)}{\left(1-\left(1+h\right)^{1/2}\right)\left(1-\left(1+h\right)^{1/3}\right)}$

Express the square root and the cube root as their binomial expansions:

$\displaystyle1-\left(1+h\right)^{1/2}=1-\left(1+\frac{h}{2}-\frac{h^2}{8}\right)=-\frac{h}{2}+\frac{h^2}{8}$
$\displaystyle1-\left(1+h\right)^{1/3}=1-\left(1+\frac{h}{3}-\frac{h^2}{9}\right)=-\frac{h}{3}+\frac{h^2}{9}$

Plug these expansions into the above.

After you do the algebra, you'll have $h^2$ as the dominant terms in both numerator and denominator. Cancel all the $h^2$ and then the answer is $\displaystyle\frac{1}{2}$

Saying what I said above, I think her teacher don't expect her to know this either, but I'll go for this one. And thank you very much.

 

[lookagain]

Express the square root and the cube root as their binomial expansions:

$\displaystyle1-\left(1+h\right)^{1/2} =1-\left(1+\frac{h}{2}-\frac{h^2}{8}\right)=...$
$\displaystyle1-\left(1+h\right)^{1/3} = 1-\left(1+\frac{h}{3}-\frac{h^2}{9}\right)=...$

Those expansions are approximate.

$\displaystyle1-\left(1+h\right)^{1/2} \approx1-\left(1+\frac{h}{2}-\frac{h^2}{8}\right)=...$
$\displaystyle1-\left(1+h\right)^{1/3}\approx 1-\left(1+\frac{h}{3}-\frac{h^2}{9}\right)=...$

Also, I don't recommend that way, nor should you give the numerical answer at this point.

 

[lookagain]

It can't be solve with L'hopital and I have no clue on how to do it.

View attachment 35186

juanpnav, you have a square root term and a cube root term. This problem lends itself to an optional substitution.

Let $\displaystyle \ u = x^6. \ \$ That will get rid of the radicals. As x approaches 1, u also approaches 1.

The expression of which you are taking the limit becomes:

$\displaystyle \dfrac{3}{1 - u^3} \ - \ \dfrac{2}{1 - u^2} \ =$

$\displaystyle \dfrac{1}{1 - u} \bigg(\dfrac{3}{1 + u + u^2} \ - \ \dfrac{2}{1 + u}\bigg) \ =$

Please continue on and post some work on the thread as needed.