# Question about Limits involving infinity

#### 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?

#### chisigma

MHF Hall of Honor
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

$$\displaystyle \chi$$ $$\displaystyle \sigma$$

#### drumist

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.

• jem163