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

Pontifex

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

(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
 
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:

\(\displaystyle \lim_{x \to \infty} \frac{1}{\sqrt{9 + \frac{1}{x}}+3}\)
 
Pontifex said:
\(\displaystyle \buildrel Lim\over{x\to\infty} \sqrt{9x^2 + x} - 3x\)

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 :
\(\displaystyle 1 = \buildrel lim\over{x\to\infty} 1\)
\(\displaystyle = \buildrel lim\over{x\to\infty} (x+1-x)\)
\(\displaystyle = \buildrel lim\over{x\to\infty} (x+1)-\buildrel lim\over{x\to\infty} x\)
\(\displaystyle = (\buildrel lim\over{x\to\infty} (x+1)-\buildrel lim\over{x\to\infty} x)*(1/x)\)
\(\displaystyle = \buildrel lim\over{x\to\infty} 1 + 1/x + \buildrel Lim\over{x\to\infty} 1\)
\(\displaystyle = 1 + 0 - 1\)
\(\displaystyle = 0\)
 
BigGlenntheHeavy said:
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 \(\displaystyle L = \frac{b}{2d}\)
 
DrMike said:
Pontifex said:
\(\displaystyle \buildrel Lim\over{x\to\infty} \sqrt{9x^2 + x} - 3x\)

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. >_>
 
daon said:
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!
 
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

\(\displaystyle Limit F(X) = 0\)
 
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