• May 28th 2010, 07:09 AM
jem163
The problem is: evaluate the limit as x approaches infinity of [x-sqrt(x)]. What I did was multiply that by [x+sqrt(x)]/[x+sqrt(x)] to get (x^2-x)/[x+sqrt(x)], i then factored an x out of the top, getting [x(x-1)]/sqrt(x). Then, since the degree of the numerator is higher, I said that the limit is infinity. I know that I got the right answer, but did I do the problem in the correct way?
• May 28th 2010, 07:32 AM
chisigma
Is...

$x - \sqrt{x} = \sqrt{x} (\sqrt{x}-1) = \frac{\sqrt{x}}{\sqrt{x}+1} (x-1)$ (1)

... and the You have to valuate the limit for $x \rightarrow \infty$ of (1)...

Kind regards

$\chi$ $\sigma$
• May 28th 2010, 08:48 AM
drumist
Quote:

Originally Posted by jem163
The problem is: evaluate the limit as x approaches infinity of [x-sqrt(x)]. What I did was multiply that by [x+sqrt(x)]/[x+sqrt(x)] to get (x^2-x)/[x+sqrt(x)], i then factored an x out of the top, getting [x(x-1)]/sqrt(x). Then, since the degree of the numerator is higher, I said that the limit is infinity. I know that I got the right answer, but did I do the problem in the correct way?

Your calculations are valid, although most of the work is unnecessary to be honest. You can consider the original expression as

$\frac{x-x^{1/2}}{1}$

Clearly already the numerator is higher degree than the denominator, and the $x$ term is the term of highest degree in the numerator, so it defines the end-behavior.