If x goes very large, say infinite, what does this equation will look like

• Feb 28th 2011, 07:01 AM
oxxiissiixxo
If x goes very large, say infinite, what does this equation will look like
when x goes very large, say infinite, what does this equation will look like?

H is a constant

1) 1/(1+(x/H)^2)^(1/2))

2) 1/(1+(H/x)^2)^(1/2))

1) 1/x^2

2) 1/x

But I am not too sure how to get there. Thank you
• Feb 28th 2011, 07:36 AM
FernandoRevilla
Quote:

Originally Posted by oxxiissiixxo
when x goes very large, say infinite, what does this equation will look like?

H is a constant

1) 1/(1+(x/H)^2)^(1/2))

2) 1/(1+(H/x)^2)^(1/2))

1) 1/x^2

2) 1/x

But I am not too sure how to get there. Thank you

Did you mean?:

$\displaystyle \dfrac{1}{\sqrt{1+(x/H)^2}}$ for 1) ?
• Feb 28th 2011, 07:42 AM
NOX Andrew
1) $\displaystyle \displaystyle \lim_{x\to\infty} \frac{1}{\sqrt{1+\left(\frac{x}{H}\right)^2}}$

As x grows larger, the denominator of the fraction grows larger. In this regard, the limit is similar to: $\displaystyle \displaystyle \lim_{t\to\infty}\frac{1}{t}$, which is 0.

2) $\displaystyle \displaystyle \lim_{x\to\infty} \frac{1}{\sqrt{1+\left(\frac{H}{x}\right)^2}}$

As x grows larger, $\displaystyle \frac{H}{x}$ approaches 0. Therefore,

$\displaystyle \displaystyle \lim_{x\to\infty} \frac{1}{\sqrt{1+\left(\frac{H}{x}\right)^2}} = \lim_{x\to\infty} \frac{1}{\sqrt{1+\left(0\right)^2}} = \lim_{x\to\infty} \frac{1}{\sqrt{1+0}} = \lim_{x\to\infty} \frac{1}{\sqrt{1}} = \lim_{x\to\infty} \frac{1}{1} = \lim_{x\to\infty} 1 = 1$