MHB Limit of (n^2+n)^(1/2)-(n^3+n^2)^(1/3)

[karseme]

How could I calculate:

$\displaystyle \lim_{n \rightarrow + \infty}{(\sqrt(n^2+n) - \sqrt[3](n^3+n^2))}$

Everything that I tried I always got infinite forms or 0 in denominator. I don't have any idea what else should I try here.

 

[MarkFL]

I would rewrite the limit as:

$$L=\lim_{n\to\infty}\left(\frac{\left(1+\dfrac{1}{n}\right)^{\frac{1}{6}}-1}{\left(n(n+1)\right)^{-\frac{1}{3}}}\right)$$

Now we have the indeterminate form 0/0 and L'Hôpital's Rule may be used. :D

 

[MarkFL]

A better approach would be to write:

$$\lim_{u\to0}\left(\frac{(u+1)^{\frac{1}{2}}-(u+1)^{\frac{1}{3}}}{u}\right)$$

Now apply L'Hôpital's Rule. ;)

 

[karseme]

Thank you on your effort, but I don't know what L'Hôpital's Rule is. I heard about it, but never used it. And we still haven't done it on my lectures. I will look it up.

 

[MarkFL]

Okay, let's see if we can do this without L'Hôpital's Rule.

Let $v=u+1$ and write the limit as:

$$L=\lim_{v\to1}\left(\frac{v^{\frac{1}{2}}-v^{\frac{1}{3}}}{v-1}\right)$$

Now, let's try to rewrite the expression so that it is not an indeterminate form:

$$\frac{v^{\frac{1}{2}}-v^{\frac{1}{3}}}{v-1}\cdot\frac{v^{\frac{1}{2}}+v^{\frac{2}{3}}}{v^{\frac{1}{2}}+v^{\frac{2}{3}}}=\frac{v^{\frac{5}{6}}\left(v^{\frac{1}{3}}-1\right)}{(v-1)\left(v^{\frac{1}{2}}+v^{\frac{2}{3}}\right)}$$

Next, let's observe that $v-1$ may be written as the difference of cubes and then factored:

$$v-1=\left(v^{\frac{1}{3}}\right)^3-1^3=\left(v^{\frac{1}{3}}-1\right)\left(v^{\frac{2}{3}}+v^{\frac{1}{3}}+1\right)$$

Hence, we have:

$$\frac{v^{\frac{1}{2}}-v^{\frac{1}{3}}}{v-1}=\frac{v^{\frac{5}{6}}}{\left(v^{\frac{2}{3}}+v^{\frac{1}{3}}+1\right)\left(v^{\frac{1}{2}}+v^{\frac{2}{3}}\right)}=\frac{v^{\frac{1}{3}}}{\left(v^{\frac{2}{3}}+v^{\frac{1}{3}}+1\right)\left(v^{\frac{1}{6}}+1\right)}$$

And thus, our limit is:

$$L=\lim_{v\to1}\left(\frac{v^{\frac{1}{3}}}{\left(v^{\frac{2}{3}}+v^{\frac{1}{3}}+1\right)\left(v^{\frac{1}{6}}+1\right)}\right)$$

Now we no longer have an indeterminate form, and we may evaluate the limit directly. :D

edit: I've moved this thread from our Analysis forum to our Calculus forum since that's a better fit for the thread topic. :D