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

Answer must be 1/4

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- May 31st 2017, 11:43 AM #1

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- May 31st 2017, 12:47 PM #2

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## 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.

- May 31st 2017, 01:31 PM #3

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- Jun 1st 2017, 02:27 PM #4

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## Re: limit3

It's not the only way. Your fraction isn't that far off the idea, either.

can be estimated to the linear form of by rewriting the radicand as

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

Likewise, can be estimated to the linear form of

Then the problem becomes: