Limits at infinity

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jtw2e2

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Having a little bit of trouble here:

Find the limit:

lim (x+2)/(?(9x^2 +1))
x->?

At first I tried to rationalize the denominator, which led to me seeing the trend of ((?)(?)) /?, so I thought the limit was ?. But looking at the graph the limit appears to be 1/3. Please help...

signed,
? confused
 
jtw2e2 said:
Having a little bit of trouble here:

Find the limit:

lim (x+2)/(?(9x^2 +1))
x->?

At first I tried to rationalize the denominator, which led to me seeing the trend of ((?)(?)) /?, so I thought the limit was ?. But looking at the graph the limit appears to be 1/3. Please help...

signed,
? confused
Click to expand...
\(\displaystyle \frac{x+2}{\sqrt{9x^2+1}} = \frac{1+\frac{2}{x}}{\sqrt{9+\frac{1}{x^2}}}\)

Do you see it now....
 
How far did you simplify the rationalized result?

Try dividing top and bottom by x, using sqrt(x^2) on the bottom.

Hold on -- I will type some LaTex formatting to show what I mean.
 
\(\displaystyle \lim_{x\to\infty}\frac{x+2}{\sqrt(9x^{2}+1)} \ = \ \lim_{x\to\infty}\frac{x+2}{\sqrt(9x^{2})}, \ 1 \ in \ the \ denominator \ is \ superfluous.\)

\(\displaystyle = \ \lim_{x\to\infty}\frac{x+2}{3x} \ = \ (Marqui) \ \lim_{x\to\infty} \frac{1}{3} \ = \ \frac{1}{3}.\)
 
Subhotosh Khan said:
\(\displaystyle \frac{x+2}{\sqrt{9x^2+1}} = \frac{1+\frac{2}{x}}{\sqrt{9+\frac{1}{x^2}}}\)

Do you see it now....
Click to expand...

Yes, thank you. It becomes 1/?9 = 1/3. So instead of trying to divide all factors by x^2, I should have recognized ?x^2 is x, and divide all factors by x?
 
You should have recognized that there is more than one way "to skin a cat"; these are but two different methods.
 
\(\displaystyle jtw2e2, \ further \ reply \ to \ your \ inquire. \ One \ of \ the \ few \ exceptions, \ when \ dealing \ with \ an \ equation,\)

\(\displaystyle is \ you \ can \ always \ multiply \ one \ side \ by \ ONE \ without \ having \ to \ do \ the \ same \ to \ the \ other \ side.\)

\(\displaystyle Hence, \ f(x) \ = \ \frac{x+2}{\sqrt(9x^{2}+1)}*\frac{1/x}{\sqrt(1/ x^{2})} \ = \ \frac{1+2/x}{\sqrt(9+1/x^{2})}\)

\(\displaystyle Now, \ when \ x \ approaches \ infinity, \ we \ are \ left \ with \ \frac{1}{3}.\)

\(\displaystyle Note: \ \lim_{x\to\infty}\frac{x+1000000000000000}{\sqrt{9x^{2}+10000000000000}} \ = \ \frac{1}{3}\)
 
jtw2e2 said:
So instead of trying to divide all factors by x^2, I should have recognized ?x^2 is x, and divide all factors by x?
Click to expand...

Yes, but I would not call them factors. We're dividing terms.

On the top, divide each term by x.

On the bottom, we divide by x, too, but because we're working inside the square-root symbol, we divide each term x^2.

\(\displaystyle \frac{\sqrt{9x^2 + 1}}{x}\)

\(\displaystyle \frac{\sqrt{9x^2 + 1}}{\sqrt{x^2}}\)

\(\displaystyle \sqrt{\frac{9x^2 + 1}{x^2}}\)

\(\displaystyle \sqrt{9 + \frac{1}{x^2}\)

This shows why the constant term 1 in the original randicand is insignificant, when x becomes infinite.
 
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