What is $\lim_{x \rightarrow \infty} \left( \sqrt{x^{2}+5x} - x \right)$?

Question

I am being asked to work out: $$\lim_{x \rightarrow \infty} \Big( \sqrt{x^{2}+5x} - x \Big)$$

My thinking is -- OK, so $x$ is some very big positive No, so I can rewrite as $$\lim_{x \rightarrow \infty} \Big( \sqrt{x^{2}+5x} - \sqrt{x^{2}} \Big)$$

I can see that $5x$ will ensure that I approach infinity.

However the actual answer is to use Taylor series approximation around zero (which in itself seems questionable, considering $x$ is approaching infinity) to arrive at an answer of $2.5$.

Where am I going wrong in my thinking. Am I?

Answer 1

You expand around $0$ by substituting $x = \frac{1}{h}$.

For a different approach, note that the limit is equal to $$\lim_{h \to 0} \frac{1}{h} \left(\sqrt{1+5h}-1\right)$$ which is manifestly $f'(0)$ where $f(x) = \sqrt{1+5x}$.

By the chain rule, $f'(x) = \frac{5}{2 \sqrt{1+5x}}$, so $f'(0) = \frac{5}{2}$.

Answer 2

Hint: A different way is to rationalise the numerator: $$\sqrt{x^2+5x}-x=\dfrac{5x}{\sqrt{x^2+5x}+x}=\dfrac{5}{\sqrt{1+\frac{5}{x}}+1}$$

Answer 3

$$\lim_{x \rightarrow \infty} \Big( \sqrt{x^{2}+5x} - x \Big)=\lim_{x \rightarrow \infty} \frac{5x}{\sqrt{x^{2}+5x}+x}.$$ Now, if you mean $x\rightarrow\infty$ as $x\rightarrow+\infty$, we obtain: $$\lim_{x \rightarrow \infty} \frac{5x}{\sqrt{x^{2}+5x}+x}=\lim_{x \rightarrow \infty} \frac{5}{\sqrt{1+\frac{5}{x}}+1}=\frac{5}{1+1}=\frac{5}{2}.$$ Also, if you mean $x\rightarrow\infty$ as $x\rightarrow\pm\infty$, so the limit does not exist because $$\lim_{x-\rightarrow-\infty}(\sqrt{x^{2}+5x}-x)=+\infty.$$

Answer 4

You have $$\sqrt{x^2 + 5x} - x = x \left(\sqrt{1 + \frac{5}{x}} - 1 \right) = \frac{5}{2} + o\left( 1\right)$$

So the limit is $\frac{5}{2}$.

Answer 5

In general,

$\begin{array}\\ \sqrt{x^2+ax}-x &=(\sqrt{x^2+ax}-x)\dfrac{\sqrt{x^2+ax}+x}{\sqrt{x^2+ax}+x}\\ &=\dfrac{x^2+ax-x^2}{\sqrt{x^2+ax}+x}\\ &=\dfrac{ax}{\sqrt{x^2+ax}+x}\\ &\to\dfrac{ax}{2x}\\ &=\dfrac{a}{2}\\ \end{array} $

Alternative proof.

$\begin{array}\\ (x+a/2)^2 &=x^2+ax+a^2/4\\ &\gt x^2+ax\\ \end{array} $

so $x+a/2 \gt \sqrt{x^2+ax} $.

$\begin{array}\\ (x+a/2-a^2/8x)^2 &=x^2+ax-a^3/(8x)+a^4/(64x^2)\\ &=x^2+ax-(a^3/(8x))(1-a/(8x))\\ &\lt x^2+ax \qquad\text{for } x > a/8\\ \end{array} $

Therefore, for $x > a/8$,

so $x+a/2 \gt \sqrt{x^2+ax} \gt x+a/2-a^2/8x $

or $0 \gt \sqrt{x^2+ax}-(x+a/2) \gt -a^2/8x $.