Question

Look at comments for question. :)

Answer 1

\[\lim_{x\rightarrow\infty}\frac{{}x^{3}2^{x}+x+1}{x^{2}3^{x}+\sqrt{2x}}\]

Answer 2

can u use L.Hopital's rule ?

Answer 3

Nope! :D

Answer 4

Initial thoughts are to divide with the dominant term 3^x and then utilize logarithm laws; I don't see where to go from there though.

Answer 5

You can place bounds on your function:
\[\frac{x^32^x}{x^23^x+\sqrt{2x}}<\frac{x^32^x+x+1}{x^23^x+\sqrt{2x}}<\frac{x^32^x+x+1}{x^23^x}\]
The leftmost expression can be simplified into
\[\frac{x(2/3)^x}{1+\sqrt{2x}/(x^23^x)}=\frac{x(2/3)^x}{1+\sqrt{2}/(x^{1/2}3^x)}\]
Thus, if you can find
\[\lim_{x\to\infty} x\left(\frac{2}{3}\right)^x\]
w/o L'Hospital, then you can use the Squeeze Theorem with the above inequality.

Answer 6

How would you find that last limit? It seems quite hard.

Answer 7

If you can find an upper bound for \(x(2/3)^x\) which goes to 0 as x goes to infinity, then that will help, because I found a lower bound for it, and that is \((2/3)^x\), which goes to 0.

Answer 8

I don't follow. :/

Answer 9

Basically, we are trying to use the Squeeze Theorem again.
If we can find some function f(x) such that \(\lim_{x\to\infty}f(x)=0\) and
\[x\left(\frac{2}{3}\right)^x<f(x)\]
then we can bound our problem by
\[\left(\frac{2}{3}\right)^x<x\left(\frac{2}{3}\right)^x<f(x)\]
And since \(\lim_{x\to\infty}\left(\frac{2}{3}\right)^x=0\), then we can conclude that \(\lim_{x\to\infty}x\left(\frac{2}{3}\right)^x=0\).

Answer 10

Had to read up on the squeeze theorem. That's a fantastic solution dude.

Answer 11

I feel bad for even remotely doubting the use of the Squeeze Theorem; never had to apply it in problems until now. This also opened my eyes on how useful it actually is, not only in theory, but also to given problems; super big thanks! :DDD

Answer 12

No problem. =)