Evaluate $\lim _{x\to \infty }\left(\sqrt{4x^2+x}-\sqrt{4x^2-1}\right)$

Question

The task is: Evaluate $\lim _{x\to \infty }\left(\sqrt{4x^2+x}-\sqrt{4x^2-1}\right)$ Can it be solved without using L'Hospital's rule? If so, what should I use?

Answer 1

Try multiplying by $\frac{\sqrt{4x^2 + x} + \sqrt{4x^2 - 1}}{\sqrt{4x^2 + x} + \sqrt{4x^2 - 1}}$. Then split into a sum of two terms and see where you get.

Answer 2

$$\sqrt{4x^2+x}-\sqrt{4x^2-1}=\frac{(4x^2+x)-(4x^2-1)}{\sqrt{4x^2+x}+\sqrt{4x^2-1}}=\frac{x+O(1)}{4x+O(1)}$$ hence the limit is $\frac{1}{4}$.

Answer 3

Multiplying numerator and denominator by the conjugate of the numerator gives $$\begin{align} \left(\sqrt{4x^2+x}-\sqrt{4x^2-1}\right)\cdot \frac{\sqrt{4x^2 + x} + \sqrt{4x^2 - 1}}{\sqrt{4x^2 + x} + \sqrt{4x^2 - 1}} & = \frac{(4x^2+x)-(4x^2-1)}{\sqrt{4x^2+x}+\sqrt{4x^2-1}} \\ \\ & = \dfrac{x+1}{\sqrt{4x^2+x}+\sqrt{4x^2-1}}\end{align}$$

Divide numerator and denominator by $x$ to get $$\lim_{x\to \infty} \dfrac{x+1}{\sqrt{4x^2+x}+\sqrt{4x^2-1}} = \lim_{x\to \infty}\dfrac{1+\frac 1x}{\sqrt {4 + \frac 1x}+ \sqrt{4 - \frac 1{x^2}}} = \frac 14$$

Answer 4

Setting $x=\dfrac1h,$

$$\lim_{x\to\infty}\sqrt{4x^2+x}-\sqrt{4x^2-1}=\lim_{h\to0^+}\left(\sqrt{\frac{4+h}{h^2}}-\sqrt{\frac{4-h^2}{h^2}}\right)$$

As $h>0,$ this reduces to $\displaystyle\lim_{h\to0^+}\frac{\sqrt{4+h}-\sqrt{4-h^2}}h$

$$\lim_{h\to0^+}\frac{\sqrt{4+h}-\sqrt{4-h^2}}h$$ $$=\lim_{h\to0^+}\frac{4+h-(4-h^2)}{h(\sqrt{4+h}+\sqrt{4-h^2})}=\lim_{h\to0^+}\frac{1+h}{\sqrt{4+h}+\sqrt{4-h^2}}=\cdots$$