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I have a quadratic equation of real $x$,

$$ x^2 - 4(1+2y)x + 8(y+1) = 0 $$

for $ x>0, y>0$ and the solution is

$$ x(y) = 4y + 2 - \sqrt{4(1+2y)^2 - 8(y+1)} $$ $$ = 4y + 2 - 2\sqrt{4y^2 + 2y -1} $$

I found the solution approaches to 1 for large $y$, if I plot $x$ vs $y$.

How can I show the solution is going to 1, for large $y$?

Want to show $$ \lim_{y \rightarrow \infty} x(y) = 1 $$

    Matlab script :
    y = linspace(0.00001, 500, 1000);
    xy = 4*y +2 - 2*(4*y.^2 + 2*y - 1).^(1/2);
    plot(y, xy)
ben Heo
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4 Answers4

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HINT:Write $$4y+2-2\sqrt{4y^2+2y-1}=2(2y+1-\sqrt{4y^2+2y-1})\frac{2y+1+\sqrt{4y^2+2y-1}}{2y+1+ \sqrt{4y^2+2y-1}}=$$ $$=2\frac{4y^2+4y+1-(4y^2+2y+1)}{2y+1+\sqrt{4y^2+2y-1}}=\frac{4y}{2y+1+\sqrt{4y^2+2y-1}}=\frac{4}{2+\frac{1}{y}+\sqrt{4+2\frac{1}{y}+\frac{1}{y^2}}}$$

Adi Dani
  • 16,949
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When y goes to infinity: $\sqrt{4y^2+2y+1}\approx \sqrt{4y^2+2y+1/4}$ so

$$4y+2-2\sqrt{4y^2+2y+1}\approx 4y+2-2\sqrt{4y^2+2y+1/4}=4y+2-2(2y+1/2)=1$$

Eli Elizirov
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$$\begin{align*} 4y+2-2\sqrt{4y^2+2y-1}&=y\left(4+\frac2y-2\sqrt{4+\frac2y-\frac1{y^2}}\right)\\\\ &=y\left(4+\frac2y-2\sqrt{4+\frac2y-\frac1{y^2}}\right)\cdot\frac{4+\frac2y+2\sqrt{4+\frac2y-\frac1{y^2}}}{4+\frac2y+2\sqrt{4+\frac2y-\frac1{y^2}}}\\\\ &=\frac{y\left(\left(4+\frac2y\right)^2-4\left(4+\frac2y-\frac1{y^2}\right)\right)}{4+\frac2y+2\sqrt{4+\frac2y-\frac1{y^2}}}\\\\ &=\frac{8+\frac4y}{4+\frac2y+2\sqrt{4+\frac2y-\frac1{y^2}}}\;; \end{align*}$$

now just take the limit as $y\to\infty$.

Brian M. Scott
  • 616,228
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$\begin{align} \sqrt{4y^2 + 2y -1} &=\sqrt{4y^2 + 4y+1-2y -2}\\ &=\sqrt{(2y+1)^2-2y -2}\\ &=\sqrt{(2y+1)^2-(2y+1) -1}\\ &=(2y+1)\sqrt{1-\frac1{2y+1} -\frac{1}{(2y+1)^2}}\\ \end{align} $

Let $z = 2y+1$. Then since $\sqrt{1+w} =1+\frac{w}{2}+O(w^2) $ as $w \to 0$,

$\begin{align} 4y + 2 - 2\sqrt{4y^2 + 2y -1} &= 2z-2z\sqrt{1-\frac1{z} -\frac{1}{z^2}}\\ &= 2z-2z(1-\frac12(\frac1{z} -\frac{1}{z^2})+O(\frac1{z^2})\\ &=2z-(2z-1+O(\frac1{z})\\ &=1+O(\frac1{z})\\ \end{align} $

Therefore $\lim_{y \to \infty} 4y + 2 - 2\sqrt{4y^2 + 2y -1} = 1 $.

marty cohen
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