Limit of $\sqrt{25x^{2}+5x}-5x$ as $x\to\infty$ $\hspace{1cm} \displaystyle\lim_{x\to\infty} \left(\sqrt{25x^{2}+5x}-5x\right) $
The correct answer seems to be $\frac12$, whereas I get $0$.
Here's how I do this problem:
$$ \sqrt{25x^{2}+5x}-5x \cdot \frac{\sqrt{25x^{2}+5x}+5x}{\sqrt{25x^{2}+5x}+5x} = \frac{25x^2+5x - 25x^2}{\sqrt{25x^{2}+5x} +5x} = \frac {5x}{\sqrt{25x^{2}+5x}+5x} $$
$\sqrt{25x^{2}+5x}$ yields a bigger value than $5x$ as $x$ becomes a very big number. So the denominator is clearly bigger than numerator. So in this case, shouldn't the answer be $0$?
However, if I keep going and divide both numerator and denominator by $x$ I get:
$$ \frac{5}{  \frac{\sqrt{25x^2+5x}}{x} + 5 }$$
In the denominator, $\frac{\sqrt{25x^2+5x}}{x}$
 yields a big number (because top is increasing faster than the bottom), in fact, it goes to infinity as $x$ goes to infinity.
In that case, it's just $5$ divided by something going to infinity, therefore, the answer should be $0$, but it's not, why?

\begin{align}
\lim_{x\to\infty} \left(\sqrt{25x^{2}+5x}-5x\right)
&= \lim_{x\to\infty} x\left(\sqrt{25+5/x}-5\right) \\
&= \lim_{x\to\infty} \frac{\sqrt{25+5/x}-5}{1/x}
\end{align}
now we can apply L'Hospital's rule:
\begin{align}
= \lim_{x\to\infty}
\left.\frac{-5}{2x^2\sqrt{25+5/x} } 
\right/
\frac{-1}{x^2}
&=
\lim_{x\to\infty}
\frac{5}{2\sqrt{25+5/x}}=\frac{1}{2}.
\end{align}
