# Thread: How do I calculate the limit of the function below?

1. ## How do I calculate the limit of the function below?

Hello, this is my first post. I hope there's someone who can help me calculate the limit of the attached function.

2. Originally Posted by azarue
Hello, this is my first post. I hope there's someone who can help me calculate the limit of the attached function.

Dear azarue,

Multiply both the numerator and denominator by, $\sqrt{4x^2+4x+1}+\sqrt{4x^2-x}$

3. ## Here's an answer I got to my private mailbox from Soroban

I can't send PM's so Thank you Soroban.

One comment, I do think there's a small mistake.
$\frac{2x+1- \sqrt{4x^2-x}}{1}\cdot\frac{2x+1 + \sqrt{4x^2-x}}{2x+1 + \sqrt{4x^2-x}} \;\;=\;\;\frac{(2x+1)^2 - (4x^2 - x)}{2x+1 + \sqrt{4x^2-x}}$ . $=\;\;\frac{3x+1}{2x+1 + \sqrt{4x^2-x}}$

The final result should be $\;\frac{5x+1}{2x+1 + \sqrt{4x^2-x}}$

Thus the limit is $\lim_{x\to\infty}=\;\;\frac{5}{4}$

Thank you all the helpers for helping me solve this.

Welcome aboard!

$\lim_{x\to\infty}\left(\sqrt{4x^2 + 4x + 1} - \sqrt{4x^2 - x}\right)$
We have: . $\sqrt{(2x+1)^2} - \sqrt{4x^2-x}\;=\;2x+1 - \sqrt{4x^2-x}$

Multiply "top and bottom" by the conjugate:

. . $\frac{2x+1- \sqrt{4x^2-x}}{1}\cdot\frac{2x+1 + \sqrt{4x^2-x}}{2x+1 + \sqrt{4x^2-x}} \;\;=\;\;\frac{(2x+1)^2 - (4x^2 - x)}{2x+1 + \sqrt{4x^2-x}}$ . $=\;\;\frac{3x+1}{2x+1 + \sqrt{4x^2-x}}$

$\text{Divide top and bottom by }x\!:\;\;\frac{\dfrac{3x}{x} + \dfrac{1}{x}}{\dfrac{2x}{x} + \dfrac{1}{x} + \dfrac{\sqrt{4x^2-x}}{x}} \;\;=\;\;\frac{3+\dfrac{1}{x}}{2 + \dfrac{1}{x} + \sqrt{\dfrac{4x^2-x}{x^2}}}$

. . . . . $=\;\;\frac{3+\dfrac{1}{x}}{2 + \dfrac{1}{x} + \sqrt{\dfrac{4x^2}{x^2} - \dfrac{1}{x^2}}} \;\;=\;\;\frac{3+\dfrac{1}{x}}{2+\dfrac{1}{x} + \sqrt{4 - \dfrac{1}{x^2}}}$

Therefore: . $\lim_{x\to\infty}\left[\frac{3+\dfrac{1}{x}}{2+\dfrac{1}{x} + \sqrt{4 - \dfrac{1}{x^2}}}\right] \;\;=\;\;\frac{3+0}{2+0+\sqrt{4-0}}\;\;=\;\;\frac{3}{4}$

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4. Originally Posted by azarue
I can't send PM's so Thank you Soroban.

One comment, I do think there's a small mistake.
$\frac{2x+1- \sqrt{4x^2-x}}{1}\cdot\frac{2x+1 + \sqrt{4x^2-x}}{2x+1 + \sqrt{4x^2-x}} \;\;=\;\;\frac{(2x+1)^2 - (4x^2 - x)}{2x+1 + \sqrt{4x^2-x}}$ . $=\;\;\frac{3x+1}{2x+1 + \sqrt{4x^2-x}}$

The final result should be $\;\frac{5x+1}{2x+1 + \sqrt{4x^2-x}}$

Thus the limit is $\lim_{x\to\infty}=\;\;\frac{5}{4}$

Thank you all the helpers for helping me solve this.

Welcome aboard!

We have: . $\sqrt{(2x+1)^2} - \sqrt{4x^2-x}\;=\;2x+1 - \sqrt{4x^2-x}$

Multiply "top and bottom" by the conjugate:

. . $\frac{2x+1- \sqrt{4x^2-x}}{1}\cdot\frac{2x+1 + \sqrt{4x^2-x}}{2x+1 + \sqrt{4x^2-x}} \;\;=\;\;\frac{(2x+1)^2 - (4x^2 - x)}{2x+1 + \sqrt{4x^2-x}}$ . $=\;\;\frac{3x+1}{2x+1 + \sqrt{4x^2-x}}$

$\text{Divide top and bottom by }x\!:\;\;\frac{\dfrac{3x}{x} + \dfrac{1}{x}}{\dfrac{2x}{x} + \dfrac{1}{x} + \dfrac{\sqrt{4x^2-x}}{x}} \;\;=\;\;\frac{3+\dfrac{1}{x}}{2 + \dfrac{1}{x} + \sqrt{\dfrac{4x^2-x}{x^2}}}$

. . . . . $=\;\;\frac{3+\dfrac{1}{x}}{2 + \dfrac{1}{x} + \sqrt{\dfrac{4x^2}{x^2} - \dfrac{1}{x^2}}} \;\;=\;\;\frac{3+\dfrac{1}{x}}{2+\dfrac{1}{x} + \sqrt{4 - \dfrac{1}{x^2}}}$

Therefore: . $\lim_{x\to\infty}\left[\frac{3+\dfrac{1}{x}}{2+\dfrac{1}{x} + \sqrt{4 - \dfrac{1}{x^2}}}\right] \;\;=\;\;\frac{3+0}{2+0+\sqrt{4-0}}\;\;=\;\;\frac{3}{4}$

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The error I see on this part :

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

is that the numerator should be:

${(2x+1)^2 - (4x^2 - x)} = 4{x^2}+4x+1 - 4{x^2}+x = 5x+1$

this should take you to the desired conclusion

5. ## I saw that issue. thank you for the comment

There was an issue in Soroban's solution, but his way to the solution was good enough to understand the logic behind it.