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?

If we start where you left off:
$$\dfrac{5x}{\sqrt{25x^{2}+5x}+5x}$$
we can factorize the square root by $5x$:
$$\dfrac{5x}{\sqrt{5x(5x+1)}+5x}$$
Take this outside of the root, and re-factorize the denominator:
$$\dfrac{5x}{\sqrt{5x}(\sqrt{5x+1}+\sqrt{5x})}$$
Cancel by $\sqrt{5x}$:
$$\dfrac{\sqrt{5x}}{\sqrt{5x+1}+\sqrt{5x}}$$
The denominator tends to $2\sqrt{5x}$ with large $x$, so the limit is $1/2$.
