Here's the problem
limit (sqrt(x^2+5x)-x)
x->[infinity]
The answer is (according to two websites and my calculator) 5/2.
Here's the work I've done so far.
[Multiply the conjugate]
(sqrt(x^2+5x)-x)/1 * (sqrt(x^2+5x)+x)/(sqrt(x^2+5x)+x)
x^2+5x-x^2/(sqrt(x^2+5x)+x) = 5x/(sqrt(x^2+5x)+x)
and I don't know how to get farther than that. Please help!
Continuing from where you left off:Originally Posted by MisanthropicMan
$\displaystyle
\displaystyle\lim_{x\to\infty} \frac{5x}{\sqrt{x^2+5x}+x} $
Here is a nice little trick: factor out an x from the denominator:
$\displaystyle = \displaystyle\lim_{x\to\infty}
\frac{5x}{x\left(\frac{\sqrt{x^2+5x}}{x} + 1\right)} $
$\displaystyle \displaystyle\lim_{x\to\infty}
= \frac{5}{\frac{\sqrt{x^2+5x}}{x} + 1}$
$\displaystyle =\displaystyle\lim_{x\to\infty}
\frac{5}{\sqrt{\frac{x^2+5x}{x^2}} + 1} $
As everything else is constant:
$\displaystyle = \frac{5}{\sqrt{\left(\displaystyle\lim_{x\to\infty }\frac{x^2+5x}{x^2}\right)} + 1} $
I trust you can continue from here?
Or, same thing, divide both numerator and denominator by x.
$\displaystyle \displaystyle\lim_{x\to\infty}
= \frac{5}{\frac{\sqrt{x^2+5x}}{x} + 1}$
$\displaystyle =\displaystyle\lim_{x\to\infty}
\frac{5}{\sqrt{\frac{x^2+5x}{x^2}} + 1} $
As everything else is constant:
$\displaystyle = \frac{5}{\sqrt{\left(\displaystyle\lim_{x\to\infty }\frac{x^2+5x}{x^2}\right)} + 1} $
I trust you can continue from here?
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