# 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?

• $\frac {5x}{\sqrt{25x^{2}+5x}+5x} \neq \frac{5x}{\sqrt{25x^2+5x}} + \frac{5x}{5x}$ Commented Mar 25, 2015 at 7:41
• $$\frac{5x}{\sqrt{25x^{2} + 5x} + 5x} \ne \frac{5x}{\sqrt{25x^{2}} + \frac{5x}{5x}$$ Commented Mar 25, 2015 at 7:41
• @dramadeur, See math.stackexchange.com/questions/1205475/… Commented Mar 25, 2015 at 8:09
• It's not really $25x^2$, it's $\sqrt{25x^2} = 5|x|$. Commented May 11, 2015 at 19:06

$$\sqrt{25x^{2}+5x}-5x \cdot \frac{\sqrt{25x^{2}+5x}+5x}{\sqrt{25x^{2}+5x}+5x} = \frac{25x^2+5x - 25x^2}{\sqrt{25x^{2}+5x} +5x} = \dfrac{5x}{\sqrt{25x^{2}+5x} +5x}$$

Your work is fine so far. Next factor out $x$ from denominator and cancel it with numerator

$$\dfrac{5x}{\sqrt{25x^{2}+5x} +5x} = \dfrac{5x}{\sqrt{x^2(25+\frac{5}{x})} +5x} =\dfrac{5x}{x\left(\sqrt{25+\frac{5}{x}} +5\right)} = \dfrac{5}{\sqrt{25+\frac{5}{x}} +5}$$

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 $x$.

• Well, I did, the denominator goes to infinity, while the numerator is simply 5. That means, the whole expression goes to $0$. But for some weird reason, it's not the correct answer. Commented Mar 25, 2015 at 7:53
• $\frac{\sqrt{25x^2+5x}}{x}$ cannot be greater than $5$ for large $x$. Try evaluating it for few $x$ values and see what you really get Commented Mar 25, 2015 at 7:54
• @ganeshie8 The way I think about it... the top is roughly 5x, right? and the bottom is just x, that means the top will be increasing faster. Commented Mar 25, 2015 at 7:55
• You're correct about top being roughly $5x$. However the expression settles to $5$ for large $x$... Notice that $5x/x = 5$ and $5$ is a fixed value, it stays firmly at $5$ no matter what $x$ is : $$\lim\limits_{x\to\infty}\dfrac{5}{1} = 5$$ Commented Mar 25, 2015 at 7:59
• There is a mistake in my first reply : I should have said "$\frac{\sqrt{25x^2+5x}}{x}$ is around $5$..." I bet you saw that already! Commented Mar 25, 2015 at 8:05

\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}

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 $1/2$.