Infinite Limits, \sqrt{9x^2 + x} - 3x

[Pontifex]

\(\displaystyle \buildrel Lim\over{x\to\infty} \sqrt{9x^2 + x} - 3x\)

(If there's a better way to denote the limit, please let me know; I've only just started playing around with TeX.)

Steps:

\(\displaystyle \buildrel Lim\over{x\to\infty} \sqrt{9x^2 + x} + \buildrel Lim\over{x\to\infty} - 3x\)
\(\displaystyle \sqrt{\buildrel Lim\over{x\to\infty} 9x^2 + x} + \buildrel Lim\over{x\to\infty} - 3x\) [Root Law]
\(\displaystyle (\sqrt{\buildrel Lim\over{x\to\infty} 9x^2 + x} + \buildrel Lim\over{x\to\infty} - 3x)*(1/x)\)
\(\displaystyle \sqrt{\buildrel Lim\over{x\to\infty} 9 + 1/x} + \buildrel Lim\over{x\to\infty} - 3\)
$\displaystyle \sqrt{9 + 0} + - 3$ [Taking the limit]
$\displaystyle 0$

The book claims $\displaystyle 1/6$, but I'm not sure where they'd get that answer.

--Pontifex

 

[daon]

Make sure to read the theorems to make sure you're applying them correctly. Sometimes intuition is dead wrong.

Try this:

$\displaystyle \lim_{x \to \infty} \frac{\sqrt{9x^2+x} - 3x}{1} \cdot \frac{\sqrt{9x^2+x}+3x}{\sqrt{9x^2+x}+3x}$

The top will simplfy immensely.

Now consider dividing the top and bottom by $\displaystyle x=\sqrt{x^2}$. You may assume x>0 with this division because it is approaching positive infinity.

With a little more algebra you'll end up with:

$\displaystyle \lim_{x \to \infty} \frac{1}{\sqrt{9 + \frac{1}{x}}+3}$

 

[BigGlenntheHeavy]

Good show, daon, as I was stumped on this one until I observed your analysis.

 

[DrMike]

\(\displaystyle \buildrel Lim\over{x\to\infty} \sqrt{9x^2 + x} - 3x\)

Steps:

\(\displaystyle \buildrel Lim\over{x\to\infty} \sqrt{9x^2 + x} + \buildrel Lim\over{x\to\infty} - 3x\) This rule only works if the limits exist.
\(\displaystyle (\sqrt{\buildrel Lim\over{x\to\infty} 9x^2 + x} + \buildrel Lim\over{x\to\infty} - 3x)*(1/x)\) Why on earth would it be valid to multiply by 1/x ??

Try this :
\(\displaystyle 1 = \buildrel lim\over{x\to\infty} 1\)
\(\displaystyle = \buildrel lim\over{x\to\infty} (x+1-x)\)
\(\displaystyle = \buildrel lim\over{x\to\infty} (x+1)-\buildrel lim\over{x\to\infty} x\)
\(\displaystyle = (\buildrel lim\over{x\to\infty} (x+1)-\buildrel lim\over{x\to\infty} x)*(1/x)\)
\(\displaystyle = \buildrel lim\over{x\to\infty} 1 + 1/x + \buildrel Lim\over{x\to\infty} 1\)
$\displaystyle = 1 + 0 - 1$
$\displaystyle = 0$

 

[daon]

Good show, daon, as I was stumped on this one until I observed your analysis.

Thanks. I had actually started taking a different route before I saw this. Consequently this leads to the following, however unhelpful it may prove in general, closed answer:

Suppsose the following:

(1) We are taking the limit of $\displaystyle \sqrt{ax^2+bx+c} - dx$
(2) $\displaystyle d^2=a$

For this limit to be real and exist, note we must have a>0. (2) ensures this, so I do not include it as an assumption.

Then:

$\displaystyle L = \lim_{x \to \infty} \sqrt{ax^2+bx+c}-dx = \frac{b}{2d}$

..

Now, what happens if we replace (2) with: (2') $\displaystyle a>0$?

We get an infinite limit, as this method reduces the limit to:

$\displaystyle L = \lim_{x \to \infty} \frac{(a-d^2)x+b+\frac{c}{x}}{\sqrt{a+\frac{b}{x}+\frac{c}{x^2}}+d}$

If $\displaystyle a>d^2$ then $\displaystyle L = \infty$
If $\displaystyle a<d^2$ then $\displaystyle L = -\infty$
If $\displaystyle a=d^2$ then we get the above result $\displaystyle L = \frac{b}{2d}$

 

[Pontifex]

\(\displaystyle \buildrel Lim\over{x\to\infty} \sqrt{9x^2 + x} - 3x\)

Steps:

\(\displaystyle \buildrel Lim\over{x\to\infty} \sqrt{9x^2 + x} + \buildrel Lim\over{x\to\infty} - 3x\) This rule only works if the limits exist.
\(\displaystyle (\sqrt{\buildrel Lim\over{x\to\infty} 9x^2 + x} + \buildrel Lim\over{x\to\infty} - 3x)*(1/x)\) Why on earth would it be valid to multiply by 1/x ??

I can has stupid newbie mistake, from time to time. >_>

 

[Pontifex]

Make sure to read the theorems to make sure you're applying them correctly. Sometimes intuition is dead wrong.

Doh! I re-read that one, thanks for the catch. I still treat infinity as a number from time to time instead of that "always increasing" idea.

Thanks!

 

[Mr.Pacman]

Basic Rule for ifinite limits....


$\displaystyle let F(x) = Ax^n/Bx^m$


if n = m

$\displaystyle Limit F(X) = A/B$



if n < m

$\displaystyle Limit F(X) = Infinity$




if n > m

$\displaystyle Limit F(X) = 0$