1. ## limit3

2x-squareroot(4x^2+x+1)
----------------------------
x-squareroot(x^2+2x)

Find the limit when x goes to infinity.

I got
2x-2x
------=undefined
x-x

2. ## Re: limit3

\begin{align*}\displaystyle \lim_{x \to \infty} \dfrac{2x-\sqrt{4x^2 + x + 1}}{x - \sqrt{x^2+2x}} \cdot \dfrac{x+\sqrt{x^2+2x}}{x+\sqrt{x^2+2x}} \cdot \dfrac{2x+\sqrt{4x^2+x+1}}{2x+\sqrt{4x^2+x+1}} & = \displaystyle \lim_{x \to \infty} \dfrac{\left[4x^2-(4x^2+x+1)\right]\left(x+\sqrt{x^2+2x}\right)}{\left[x^2-(x^2+2x)\right]\left(2x+\sqrt{4x^2+x+1}\right)} \\ & = \displaystyle \lim_{x \to \infty} \dfrac{(x+1)\left(x+\sqrt{x^2+2x}\right)}{2x\left( 2x+\sqrt{4x^2+x+1}\right)}\cdot \dfrac{\dfrac{1}{x^2}}{\dfrac{1}{x^2}} \\ & = \displaystyle \lim_{x \to \infty} \dfrac{\left(1+\dfrac{1}{x}\right)\left(1+\sqrt{1+ \dfrac{2}{x}}\right)}{2\left(2+\sqrt{4+\dfrac{1}{x }+\dfrac{1}{x^2}}\right)} \\ & = \dfrac{(1+0)\left(1+\sqrt{1+0}\right)}{2(2+\sqrt{4 +0+0})} \\ & = \dfrac{1}{4}\end{align*}

I skipped a bunch of steps. Let me know if you have trouble following anything I did.

3. ## Re: limit3

2x-2x
------=undefined
x-x
I wanted to solve by that way but it is impossible in this example. Your solution is the only way . Many Thanks.

4. ## Re: limit3

Originally Posted by kastamonu
2x-2x
------=undefined
x-x
I wanted to solve by that way but it is impossible in this example. Your solution is the only way . Many Thanks.
It's not the only way. Your fraction isn't that far off the idea, either.

$\sqrt{4x^2 + x + 1}$ can be estimated to the linear form of $2x + \tfrac{1}{4}$ by rewriting the radicand as $(2x + \tfrac{1}{4})^2 + \tfrac{15}{16}$

by completing the square, for instance, or using the binomial theorem, for two of the ways.

Likewise, $\sqrt{x^2 + 2x}$ can be estimated to the linear form of $x + 1.$

Then the problem becomes:

$\displaystyle\lim_{x \to \infty} \dfrac{2x - (2x + \tfrac{1}{4})}{x - (x + 1)} =$

$\displaystyle\lim_{x \to \infty} \dfrac{2x - 2x - \tfrac{1}{4}}{x - x - 1} =$

$\dfrac{- (\tfrac{1}{4})}{ - 1} =$

$\dfrac{1}{4}$