Calculate $\lim_{n\to\infty} \sqrt{5n^2+4}~-~\sqrt{5n^2+n}$

Question

Calculate:$$\lim_{n\to\infty} \sqrt{5n^2+4}~-~\sqrt{5n^2+n}$$

Even though I know how to handle limits like this, I would be interested in other ways to approach to tasks similar to this. My own solution will be at the bottom.

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My own solution

$$\lim_{n\to\infty} \sqrt{5n^2+4}~-~\sqrt{5n^2+n}~\frac{\sqrt{5n^2+4}~+~\sqrt{5n^2+n}}{\sqrt{5n^2+4}~+~\sqrt{5n^2+n}}$$ $$\lim_{n\to\infty} \frac{5n^2+4~-~(5n^2+n)}{\sqrt{5n^2+4}~+~\sqrt{5n^2+n}}~=~\lim_{n\to\infty} \frac{4-n}{n\left(\sqrt{5+\frac4{n^2}}~+~\sqrt{5+\frac1{n}}\right)}$$ $$=~\frac{-1}{2\sqrt{5}}~=~-\frac{\sqrt{5}}{10}$$

Answer 1

Two alternative ideas:

(1) Write $$ \sqrt{5n^2+4} - \sqrt{5n^2+n} = \sqrt 5n(\sqrt{1 + 4/5n^2} - \sqrt{1 + 1/5n}) $$ and use Taylor ($\sqrt{1 + x} = 1 + x/2 - x^2/8 + \cdots$).

(2) Let be $f(x) = \sqrt x$. By the Mean Value Theorem, for some $c_n\in(5n^2 + 4,5n^2 + n)$ $$ \sqrt{5n^2 + 4} - \sqrt{5n^2 + n} = f'(c_n)((5n^2 + 4) - (5n^2 + n)) = -\frac{4 - n}{2\sqrt{c_n}} = -\frac{4/n - 1}{2\sqrt{c_n/n^2}}, $$ and by sqeezing $c_n/n^2\to 5$.

Answer 2

Let $1/n=h$

As $n\to\infty,h\to0+,h>0$

$$\sqrt{5n^2+4}=\sqrt{\dfrac{5+4h^2}{h^2}}=\dfrac{\sqrt{5+4h^2}}{\sqrt{h^2}}$$

Now $\sqrt{h^2}=|h|=+h$ for $h>0$

So, we ahve $$\lim_{h\to0^+}\dfrac{\sqrt{5+4h^2}-\sqrt{5+h}}h=\lim_{h\to0^+}\dfrac1{\sqrt{5+4h^2}+\sqrt{5+h}}\cdot\lim_{h\to0^+}\dfrac{5+4h^2-(5+h)}h=?$$

Answer 3

Considering that $\frac{4}{n^2}<< \frac{1}{n}$ for big $n$ we have

$$ \lim_{n\to\infty} \sqrt{5n^2+4}~-~\sqrt{5n^2+n} \equiv \lim_{h\to 0}\frac{\sqrt{5+4h^2}-\sqrt{5+h}}{h} = \lim_{h\to 0}\frac{\sqrt{5+o(h^2)}-\sqrt{5+h}}{h} $$

now remember the derivative definition

$$ \lim_{h\to 0}\frac{\sqrt{x+h}-\sqrt x}{h} = \frac 12\frac{\sqrt x}{x} $$

so the result is

$$ -\frac 12\frac{\sqrt 5}{5} $$