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?
 A: $$\sqrt{25x^{2}+5x}-5x \cdot \frac{\sqrt{25x^{2}+5x}+5x}{\sqrt{25x^{2}+5x}+5x} = \frac{25x^2+5x - 25x^2}{\sqrt{25x^{2}+5x} +5x} = \dfrac{5x}{\sqrt{25x^{2}+5x} +5x}$$
Your work is fine so far. Next factor out $x$ from denominator and cancel it with numerator
$$ \dfrac{5x}{\sqrt{25x^{2}+5x} +5x} =  \dfrac{5x}{\sqrt{x^2(25+\frac{5}{x})} +5x} =\dfrac{5x}{x\left(\sqrt{25+\frac{5}{x}} +5\right)}  = \dfrac{5}{\sqrt{25+\frac{5}{x}} +5} $$
A: First of all, the step
$$
\frac{5x}{\sqrt{25x^2+5x}+5x} =\frac{5x}{\sqrt{25x^2+5x}}+\frac{5x}{5x}
$$
is wrong, you cannot separate the fractions like that. This said, you can try the following:
$$
\frac{5x}{\sqrt{25x^2+5x}+5x} = \frac{\frac{5x}{x}}{\frac{\sqrt{25x^2+5x}+5x}{x}},
$$
that is, dividing the numerator and denominator by $x$.
A: \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}
A: 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 $1/2$.
