Find the limit.
$\displaystyle \sqrt{25x^2 + x} - 5x = \frac{(\sqrt{25x^2 + x} - 5x)(\sqrt{25x^2 + x} + 5x)}{(\sqrt{25x^2 + x} + 5x)} = \frac{25x^2 + 5x - 25x^2}{\sqrt{25x^2 + x} + 5x}$
$\displaystyle = \frac{5x}{\sqrt{25x^2 + x} + 5x} = \frac{5}{\sqrt{25 + \frac{1}{x}} + 5} .$
Take the limit as $\displaystyle x \rightarrow \infty$ of that.
To simplify $\displaystyle \frac{(\sqrt{25x^2 + x} - 5x)(\sqrt{25x^2 + x} + 5x)}{(\sqrt{25x^2 + x} + 5x)}$,
recall: $\displaystyle (A - B)(A + B) = A^2 - B^2$.
Let $\displaystyle A = \sqrt{25x^2 + x}$ and $\displaystyle B = 5x$.
Then $\displaystyle A^2 - B^2 = 25x^2 + x - (5x)^2 = 25x^2 + x - 25x^2 = x.$
Therefore $\displaystyle
\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}.
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Typo noted - thanks for that get, Jhevon.