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))
The answers are
1) 1/x^2
2) 1/x
But I am not too sure how to get there. Thank you
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))
The answers are
1) 1/x^2
2) 1/x
But I am not too sure how to get there. Thank you
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,
$\displaystyle \displaystyle \lim_{x\to\infty} \frac{1}{\sqrt{1+\left(\frac{H}{x}\right)^2}} = \lim_{x\to\infty} \frac{1}{\sqrt{1+\left(0\right)^2}} = \lim_{x\to\infty} \frac{1}{\sqrt{1+0}} = \lim_{x\to\infty} \frac{1}{\sqrt{1}} = \lim_{x\to\infty} \frac{1}{1} = \lim_{x\to\infty} 1 = 1$