L'Hopital's Rule -- using conjugate to find answer

[Guest]

For the following limit:

. . .lim_{x--> infinity} (x^3+2x+5)^(1/3) - x

Am I just suppoesd to use a conjugate to find the answer?

 

[stapel]

Am I just suppoesd to use a conjugate to find the answer?

Have you tried this? What did you get?

Please be specific. Thank you.

Eliz.

 

[galactus]

There's a trick you can use.

Rewrite \(\displaystyle \L\\\lim_{x\to\infty}(x^{3}+2x+5)^{\frac{1}{3}}-x\) as

\(\displaystyle \L\\\lim_{x\to\infty}\frac{2x+5}{x^{2}+x(x^{3}+2x+5)^{\frac{1}{3}}+(x^{3}+2x+5)^{\frac{2}{3}}}\)


\(\displaystyle \L\\\lim_{0\to\infty}\frac{2+\frac{5}{x}}{x+\sqrt{x^{3}+2x+5}+(x^{\frac{3}{2}}+\frac{2}{\sqrt{2}}+\frac{5}{x^{\frac{3}{2}}})\)

Can you see what the limit is now. I know it looks ominous.

 

[soroban]

Hello, aswimmer113!

\(\displaystyle \lim_{x\to\infty}\,\left[(x^3\,+2x\,+\,5)^{\frac{1}{3}}\,-\,x\right]\)

You're right . . . we need the conjugate.

Galactus' trick involves this conjugate . . . I'll try to explain this.


We know that: \(\displaystyle \,p^3\,-\,q^3\;=\;(p\,-\,q)(p^2\,+\,pq\,+\,q^2)\)

. . where \(\displaystyle p^2\,+\,pq\,+\,q^2\) is the conjugate of \(\displaystyle p\,-\,q.\)


Then, surprisingly, \(\displaystyle \,a\,-\,b\) can be "factored" . . . . like this:

\(\displaystyle \;\;\;a\,-\,b\;=\;\left(\sqrt[3]{a}\right)^3\,-\,\left(\sqrt[3]{b}\right)^3 \;= \;\left(\sqrt[3]{a}\,-\,\sqrt[3]{b}\right)\,\left(\sqrt[3]{a^2}\,+\,\sqrt[3]{ab}\,+\,\sqrt[3]{b^2}\right)\)


We have: \(\displaystyle \,\underbrace{\sqrt[3]{x^3\,+\,2x\,+\,5}}\,-\,\underbrace{x}\)
. . . . . . . . . . . . . . \(\displaystyle \sqrt[3]a\) . . . . . . . \(\displaystyle \sqrt[3]b\)

The conugate is: \(\displaystyle \,\left[\sqrt[3]{x^3+2x+5}\right]^2 \,+ \,\sqrt[3]{x^3+2x+5}\,\cdot\,x\,+\,x^2\)

. . . . . . . . . . . \(\displaystyle = \;(x^3\,+\,2x\,+\,5)^{\frac{1}{3}}\,+\,x(x^3+2x+5)^{\frac{1}{3}}\,+\,x^2\)