How to find an exact limit of a function which is subtracting square roots of quadratic equations?

Question

Find exact value of limit:

$$ \lim \limits_{x \to \infty} \sqrt{( 3x^2 + 8x + 6)}-\sqrt{( 3x^2+3x+4)} $$

Here is what I've got so far:

$$\lim \limits_{x \to \infty} \frac{(\sqrt{( 3x^2 + 8x + 6)}-\sqrt{( 3x^2+3x+4)})-(\sqrt{( 3x^2 + 8x + 6)}+\sqrt{( 3x^2+3x+4)})}{\sqrt{( 3x^2 + 8x + 6)}-\sqrt{( 3x^2+3x+4)}} $$

$$\lim \limits_{x \to \infty} \frac{3x^2+8x+6-3x^2-3x-4}{\sqrt{( 3x^2 + 8x + 6)}-\sqrt{( 3x^2+3x+4)}} $$

$$\lim \limits_{x \to \infty} \frac{5x+2}{\sqrt{( 3x^2 + 8x + 6)}-\sqrt{( 3x^2+3x+4)}} $$

At this point I am stuck.. Hopefully someone can help me out with this one.

Answer 1

Correcting the mistake I mentioned in the comments we have $$\lim_{x\to\infty} \frac{5x+2}{\sqrt{3x^2+8x+6} + \sqrt{3x^2+3x+4}}$$ $$=\lim_{x\to\infty} \frac{5x+2}{x\left(\sqrt{3+8/x+6/x^2} + \sqrt{3+3/x+4/x^2}\right)}$$ $$=\lim_{x\to\infty} \frac{5+2/x}{\sqrt{3+8/x+6/x^2} + \sqrt{3+3/x+4/x^2}}$$ $$=\frac{5}{2\sqrt{3}}$$

Answer 2

you have $$\lim_{x\rightarrow\infty}\sqrt{ax^{2}+bx + c} - \sqrt{px^{2}+qx+r}$$ first burn the radicals via $$\lim_{x\rightarrow\infty}\frac{(\sqrt{ax^{2}+bx + c} - \sqrt{px^{2}+qx+r})(\sqrt{ax^{2}+bx + c} + \sqrt{px^{2}+qx+r})}{(\sqrt{ax^{2}+bx + c} + \sqrt{px^{2}+qx+r})}$$ to get $$\lim_{x\rightarrow\infty}\frac{ax^{2}+bx+c-px^{2}-qx-r}{(\sqrt{ax^{2}+bx + c} + \sqrt{px^{2}+qx+r})}$$ now, notice that if $a \neq p$ then this expression tends towards $\infty$ ($a > p$) or $-\infty$ ($a < p$) since the numerator is on the order of $x^{2}$ while the denominator is on the order of $x$.

If $a = p$ then we have $$\lim_{x\rightarrow\infty}\frac{(b-q)x+c-r}{(\sqrt{ax^{2}+bx + c} + \sqrt{px^{2}+qx+r})}$$

From here there is the usual method of factoring out the greatest power of $x$ to obtain

$$\lim_{x\rightarrow\infty}\frac{(b-q)+\frac{c-r}{x}}{(\sqrt{a+\frac{b}{x} + \frac{c}{x^{2}}} + \sqrt{p+\frac{q}{x}+\frac{r}{x^{2}}})}$$

which yields $$\frac{b-q}{2\sqrt{a}}$$ since $a = p$.

In your case, $a = p = 3$, $b = 8$, $q = 3$ and so you have that the limit is $$\frac{5}{2\sqrt{3}} = \frac{5\sqrt{3}}{6}$$

Answer 3

Just to show you another way to do it $$A= \sqrt{( 3x^2 + 8x + 6)}-\sqrt{( 3x^2+3x+4)}=x \sqrt 3\left(\sqrt{1+\frac 8{3x}+\frac 2{x^2}}-\sqrt{1+\frac 1{x}+\frac 4{3x^2}} \right)$$ Now, use Taylor which, for small $y$, is $$\sqrt{1+y}=1+\frac{y}{2}-\frac{y^2}{8}+O\left(y^3\right)$$ In the first radical, replace $y$ by $(\frac 8{3x}+\frac 2{x^2})$ and in the second radical, replace $y$ by $(\frac 1{x}+\frac 4{3x^2})$.

Limiting to $O\left(\frac{1}{x^3}\right)$, you will then have $$A=x \sqrt 3\left( 1+\frac{4}{3 x}+\frac{1}{9 x^2}-\frac{4}{27 x^3}+O\left(\frac{1}{x^3}\right)-1-\frac{1}{2 x}-\frac{13}{24 x^2}+O\left(\frac{1}{x^3}\right)\right)$$ $$A=\frac{5}{2 \sqrt{3}}-\frac{31}{24 \sqrt{3} x}+O\left(\frac{1}{x^2}\right)$$ which not only shows the limit but also how it is approached.