limits

[jon12]

I need to find the limit as x approaches infinity for this problem: ?(4x^2+1)-2x I know some how I will have to conver the limit with the 'change of variables' technique involving L'Hopitals rule. Need some help. Someone show an example of how this might work in a similar situation or show me how to work this out?

 

[soroban]

Hello, jon12!

\(\displaystyle \lim_{x\to\infty} \left(\sqrt{4x^2+1} - 2x\right)\)

Multiply top and bottom by the conjugate of the numerator:

. . \(\displaystyle \frac{\sqrt{4x^2+1} - 2x}{1}\cdot \frac{\sqrt{4x^2+1} + 2x}{\sqrt{4x^2+1} + 2x} \;\;=\;\; \frac{(4x^2+1) - 4x^2}{\sqrt{4x^2+1} + 2x} \;=\;\frac{1}{\sqrt{4x^2+1}+2x}\)


Divide top and bottom by \(\displaystyle x\!:\)

. . \(\displaystyle \displaystyle \frac{\dfrac{1}{x}}{\dfrac{\sqrt{4x^2+1}}{x} + \dfrac{2x}{x}} \;\;=\;\;\frac{\dfrac{1}{x}}{\sqrt{\dfrac{4x^2+1}{x^2}} + 2} \;=\;\frac{\dfrac{1}{x}}{\sqrt{\dfrac{4x^2}{x^2} + \dfrac{1}{x^2}} + 2} \;\;=\;\;\frac{\dfrac{1}{x}}{\sqrt{4 + \dfrac{1}{x^2}} + 2}\)


\(\displaystyle \displaystyle \text{Therefore: }\;\lim_{x\to\infty} \frac{\dfrac{1}{x}}{\sqrt{4 + \dfrac{1}{x^2}} + 2} \;\;=\;\;\frac{0}{\sqrt{4+0} + 2} \;\;=\;\;\frac{0}{4} \;=\;0\)