# Homework Help: Limit of a function

1. Dec 12, 2007

### fk378

1. The problem statement, all variables and given/known data
Find the limit as x-->infinity of (x+2)/sqrt (9x^2 + 1)

2. Relevant equations
look at the dominating terms

3. The attempt at a solution
I squared the entire function to get rid of the square root (is this allowed? Or am I changing the function?) And I get (x^2 + 4x + 4)/ (9x^2 +1)
the dominating term on the top is x^2 and the dominating term on the bottom is 9x^2. Therefore, the limit as x-->infinity is x^2/9x^2 = 1/9

My professor says the answer is 1/3...

2. Dec 12, 2007

### ircdan

maybe it's easier to see if you divide the numerator and denominator by x,
(x+2)/sqrt(9x^2 + 1) = (1 + 2/x)/sqrt(9 + 1/x^2) -> 1/sqrt(9) = 1/3 as x->inf

3. Dec 12, 2007

### rock.freak667

$$lim_{x\rightarrow \infty}\frac{x+2}{\sqrt{9x^2 + 1}}$$

can be written as

$$lim_{x\rightarrow \infty}\sqrt{\frac{(x+2)^2}{{9x^2 + 1}}}$$

and the fraction works out to be (x^2+4x+4)/(9x^2+1)

so dividing by the highest power of x and taking the limit you get

$$lim_{x\rightarrow \infty}\sqrt{\frac{(x+2)^2}{{9x^2 + 1}}}=\sqrt{\frac{1}{9}}$$

so you had to take the square root of the answer you got

4. Dec 12, 2007

### fk378

Oh I see what you're both saying. But I squared the entire function, that's why I didn't take the square root of my answer at the end, because the square root was gone...am I supposed to take it anyway to undo the squaring that I did?

Also, when I try to solve it by taking the conjugate radical of the denominator, I don't get 1/3 either. In fact, I get x^2 which means the limit would be approaching infinity.

Trying L'Hospital's rule doesn't work either.

Last edited: Dec 12, 2007
5. Dec 12, 2007

### dynamicsolo

Yes.

Are you saying you would multiply numerator and denominator by sqrt[(9x^2) - 1]? It should work eventually, but it sure doesn't make things any easier...

L'Hopital's Rule will work (eventually), but it makes quite a mess of such expressions in the process... The approach already described by ircdan and rock.freak667 really is the quickest way to go.

[And what is with this "T" and "Y" for time-stamping posts? Is PF needing to save "ink"?]

6. Dec 12, 2007

### fk378

I don't see how L'Hospital would work, though. If you get the derivative of the top, it's 1, and the derivative of the bottom is (1/2)(9x^2+1)(18x), which equals 16x^3+18x....so 1/(16x^3+18x) would still be indeterminate, so you do L'Hospital again, and you get 0....is that right? (would the denominator's derivative be arctan3x instead?)

Thanks...I'm just trying to figure out why certain methods would NOT work...

7. Dec 12, 2007

### dynamicsolo

In the original expression, the derivative of the denominator would be

(1/2) · [(9x^2+1)^(-1/2)] · (18x) .

[I believe you're thinking of your squared expression, for which L'Hopital's Rule would work quite easily.]

You would now have to take the limit of

[ sqrt(9x^2+1) ]/9x .

This is still indeterminate, but if you use L'Hopital yet again, an x is again left in the numerator and the radical returns to the denominator. You could play this game of differential ping-pong indefinitely. But the technique of dividing this last limit by x in the numerator and denominator will still give [sqrt(9)]/9 = 1/3 .

This is one of those expressions which L'Hopital's Rule does not resolve. It does the situation no harm, but it doesn't get you to the answer either...