How to solve this limit
- Thread starter juanpnav
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Start by putting them over a common denominator: [imath]\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)}[/imath]
Then let [imath]\displaystyle x=\lim_{h\to0}\left(1+h\right)[/imath], and substitute in:
[imath]\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)}[/imath]
Express the square root and the cube root as their binomial expansions:
[imath]\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}[/imath]
[imath]\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}[/imath]
Plug these expansions into the above.
After you do the algebra, you'll have [imath]h^2[/imath] as the dominant terms in both numerator and denominator. Cancel all the [imath]h^2[/imath] and then the answer is [imath]\displaystyle\frac{1}{2}[/imath]
Then let [imath]\displaystyle x=\lim_{h\to0}\left(1+h\right)[/imath], and substitute in:
[imath]\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)}[/imath]
Express the square root and the cube root as their binomial expansions:
[imath]\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}[/imath]
[imath]\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}[/imath]
Plug these expansions into the above.
After you do the algebra, you'll have [imath]h^2[/imath] as the dominant terms in both numerator and denominator. Cancel all the [imath]h^2[/imath] and then the answer is [imath]\displaystyle\frac{1}{2}[/imath]
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.
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.
Those expansions are approximate.
[imath]\displaystyle1-\left(1+h\right)^{1/2} \approx1-\left(1+\frac{h}{2}-\frac{h^2}{8}\right)=...[/imath]
[imath]\displaystyle1-\left(1+h\right)^{1/3}\approx 1-\left(1+\frac{h}{3}-\frac{h^2}{9}\right)=...[/imath]
Also, I don't recommend that way, nor should you give the numerical answer at this point.
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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.