Convergence Proof: $\lim_{x\rightarrow\infty} \sqrt{4x+x^2}- \sqrt{x^2+x}$

Question

I have to check whether the following expression converges; if yes I have to give the limit.

$$\lim_{x\rightarrow\infty} \sqrt{4x+x^2}- \sqrt{x^2+x}$$

Now I did the following:

$$\lim_{x\rightarrow\infty} \sqrt{4x+x^2}- \sqrt{x^2+x}$$ $$\lim_{x\rightarrow\infty} x\sqrt{\frac{4}{x}+1}- x\sqrt{1+\frac{1}{x}}$$ $$\lim_{x\rightarrow\infty} x \lim_{x\rightarrow\infty}(\sqrt{\frac{4}{x}+1}- \sqrt{1+\frac{1}{x}})$$

That confuses me. The left limit approaches $\infty$ while the right approaches $0$. What is wrong here or what can I conclude from that?

Thank you very muvh for your help in advance!

FunkyPeanut

Answer 1

Hint: Multiply by $\frac{\sqrt{4x+x^2}+\sqrt{x^2+x}}{\sqrt{4x+x^2}+\sqrt{x^2+x}}$.

Answer 2

You get nowhere by reasoning that way. The canonical way to work on this is "rationalizing": you multiply and divide by $\sqrt{4x+x^2}+\sqrt{x^2+x}$. That way you get $$ \sqrt{4x+x^2}-\sqrt{x^2+x}=\frac{4x+x^2-x^2-x}{\sqrt{4x+x^2}+\sqrt{x^2+x}}=\frac{3x}{\sqrt{4x+x^2}+\sqrt{x^2+x}}=\frac3{\sqrt{4/x+1}+\sqrt{1+1/x}}\to\frac32 $$

Answer 3

$\lim_{x\rightarrow\infty} \sqrt{4x+x^2}- \sqrt{x^2+x} = \lim_{x\rightarrow\infty} \frac{(\sqrt{4x+x^2}- \sqrt{x^2+x})\cdot(\sqrt{4x+x^2}+ \sqrt{x^2+x})}{ \sqrt{4x+x^2}+ \sqrt{x^2+x}} = \lim_{x\rightarrow\infty} \frac{4x+x^2-x^2-x}{ \sqrt{4x+x^2}+ \sqrt{x^2+x}} = \lim_{x\rightarrow\infty} \frac{3x}{ \sqrt{4x+x^2}+ \sqrt{x^2+x}} = \lim_{x\rightarrow\infty} \frac{3}{ \sqrt{\frac{4}{x}+1}+ \sqrt{1+\frac{1}{x}}} = \frac{3}{ \lim_{x\rightarrow\infty}\sqrt{\frac{4}{x}+1}+ \sqrt{1+\frac{1}{x}}} = \frac{3}{\sqrt{1}+\sqrt{1}} =\frac{3}{2}$

Answer 4

You can also continue after having factored $x$ as you did. Now, remember that when $y$ is small
$\sqrt{y+1}$ can be approximated by $1+\frac{y}{2}$ (first term of Taylor expansion). So,
$$\sqrt{\frac{4}{x}+1}- \sqrt{1+\frac{1}{x}}=(1+\frac{2}{x})-(1+\frac{1}{2x})=\frac{3}{2x}$$ Multiplying by the $x$ you factored, you get the required limit.

Answer 5

Explanation of the trick (see previous answers) and generalization. Is based in the formula $$a-b = {(a-b)(a+b)\over a+b} = {a^2-b^2\over a+b}.$$ Applied to $\sqrt\cdots-\sqrt\cdots$, the squares kill the roots and the sum of roots in the denominator is not a problem.

With the factorization $(a^n-b^n)=(a-b)(a^{n-1}+\cdots+b^{n-1})$ is possible to eliminate a difference of $n$-th roots $\root n\of\cdots-\root n\of\cdots$.