1. Find the limit.

Find the limit.

2. Originally Posted by plstevens
Find the limit.

$\sqrt{25x^2 + x} - 5x = \frac{(\sqrt{25x^2 + x} - 5x)(\sqrt{25x^2 + x} + 5x)}{(\sqrt{25x^2 + x} + 5x)} = \frac{25x^2 + 5x - 25x^2}{\sqrt{25x^2 + x} + 5x}$

$= \frac{5x}{\sqrt{25x^2 + x} + 5x} = \frac{5}{\sqrt{25 + \frac{1}{x}} + 5} .$

Take the limit as $x \rightarrow \infty$ of that.

3. Originally Posted by mr fantastic
$\sqrt{25x^2 + x} - 5x = \frac{(\sqrt{25x^2 + x} - 5x)(\sqrt{25x^2 + x} + 5x)}{(\sqrt{25x^2 + x} + 5x)} = \frac{25x^2 + {\color{red} 5x} - 25x^2}{\sqrt{25x^2 + x} + 5x}$
...[/tex]

Take the limit as $x \rightarrow \infty$ of that.
you have a small typo mr fantastic. i was going to ask plstevens to find it, but i said never mind. the term highlighted should be $x$

4. i don't understand how u got 5x, when u multiple the top out u get 25x^2+x+5x? cause i still don't understand how u got what u got

5. Originally Posted by plstevens
i don't understand how u got 5x, when u multiple the top out u get 25x^2+x+5x? cause i still don't understand how u got what u got
To simplify $\frac{(\sqrt{25x^2 + x} - 5x)(\sqrt{25x^2 + x} + 5x)}{(\sqrt{25x^2 + x} + 5x)}$,

recall: $(A - B)(A + B) = A^2 - B^2$.

Let $A = \sqrt{25x^2 + x}$ and $B = 5x$.

Then $A^2 - B^2 = 25x^2 + x - (5x)^2 = 25x^2 + x - 25x^2 = x.$

Therefore $
\frac{(\sqrt{25x^2 + x} - 5x)(\sqrt{25x^2 + x} + 5x)}{(\sqrt{25x^2 + x} + 5x)} = \frac{25x^2 + x - 25x^2}{\sqrt{25x^2 + x} + 5x} = \frac{x}{\sqrt{25x^2 + x} + 5x}.
$

Typo noted - thanks for that get, Jhevon.