Question about Limits involving infinity

May 2010
1
0
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?
 

chisigma

MHF Hall of Honor
Mar 2009
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near Piacenza (Italy)
Is...

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

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

Kind regards

\(\displaystyle \chi\) \(\displaystyle \sigma\)
 
Jan 2010
354
173
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

\(\displaystyle \frac{x-x^{1/2}}{1}\)

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