# Find the limit of square root

• Jun 27th 2010, 10:22 PM
p75213
Find the limit of square root
The question is to find the limit as x->inf of x/sqr(1+(x/c)^2) where c is a constant. I rewrote the equation as sqr(x^2*c^2/(x^2+c^2)). Then find the limit of what is inside the square root ie. x^2*c^2/(x^2+c^2). This comes to c^2 the square root of which is c.

Is this the correct and most appropriate way of arriving at the limit?
• Jun 27th 2010, 11:08 PM
Jhevon
A bit unorthodox, but that works. you forgot one thing though: $\sqrt{c^2} = |c|$
• Jun 28th 2010, 12:07 AM
p75213
I thought it may be somewhat unorthodox. What would be a more common method?
• Jun 28th 2010, 12:27 AM
nehme007
Divide both the numerator and denominator by x. (Just x because it'll get squared when it goes under the square root and kill the x^2 under the square root in the denominator).

$\lim_{x \rightarrow \infty} \frac{x}{\sqrt{1+(x/c)^2}} = \lim_{x \rightarrow \infty} \frac{1}{\sqrt{1/x^2+(1/c)^2}} = \frac{1}{\sqrt{(1/c)^2}} = |c|$
• Jun 28th 2010, 12:57 AM
p75213
Quote:

Originally Posted by nehme007
Divide both the numerator and denominator by x. (Just x because it'll get squared when it goes under the square root and kill the x^2 under the square root in the denominator).

$\lim_{x \rightarrow \infty} \frac{x}{\sqrt{1+(x/c)^2}} = \lim_{x \rightarrow \infty} \frac{1}{\sqrt{1/x^2+(1/c)^2}} = \frac{1}{\sqrt{(1/c)^2}} = |c|$

Less algebraic manipulation required with this method and therefore quicker and better.