# Thread: Would like Help with Calc problem!

1. ## Would like Help with Calc problem!

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)

2. Originally Posted by MisanthropicMan
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)

Continuing from where you left off:
$
\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:

$= \displaystyle\lim_{x\to\infty}
\frac{5x}{x\left(\frac{\sqrt{x^2+5x}}{x} + 1\right)}$

$\displaystyle\lim_{x\to\infty}
= \frac{5}{\frac{\sqrt{x^2+5x}}{x} + 1}$

$=\displaystyle\lim_{x\to\infty}
\frac{5}{\sqrt{\frac{x^2+5x}{x^2}} + 1}$

As everything else is constant:

$= \frac{5}{\sqrt{\left(\displaystyle\lim_{x\to\infty }\frac{x^2+5x}{x^2}\right)} + 1}$

I trust you can continue from here?

3. Originally Posted by Gusbob
Continuing from where you left off:
$
\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:

$= \displaystyle\lim_{x\to\infty}
\frac{5x}{x\left(\frac{\sqrt{x^2+5x}}{x} + 1\right)}$
Or, same thing, divide both numerator and denominator by x.

$\displaystyle\lim_{x\to\infty}
= \frac{5}{\frac{\sqrt{x^2+5x}}{x} + 1}$

$=\displaystyle\lim_{x\to\infty}
\frac{5}{\sqrt{\frac{x^2+5x}{x^2}} + 1}$

As everything else is constant:

$= \frac{5}{\sqrt{\left(\displaystyle\lim_{x\to\infty }\frac{x^2+5x}{x^2}\right)} + 1}$

I trust you can continue from here?