Limits

ctran · #1 $Old$ November 6th, 2009, 09:22 PM

How would you compute this limit?

lim (3x-sqrt(9x^(2)+10))/ (3x-1000)
x-> -inf

Debsta · #2 $Old$ November 6th, 2009, 09:25 PM

I'd use l'Hopitals rule twice.

alexmahone · #3 $Old$ November 6th, 2009, 09:40 PM

Divide numerator and denominator by x.

ctran · #4 $Old$ November 6th, 2009, 10:08 PM

Quote:

Originally Posted by alexmahone $View Post$

Divide numerator and denominator by x.

Then use l'hopital's rule twice?

Debsta · #5 $Old$ November 6th, 2009, 10:11 PM

No not "then". One or the other.

ctran · #6 $Old$ November 6th, 2009, 10:18 PM

Quote:

Originally Posted by Debsta $View Post$

No not "then". One or the other.

Okay if I divide it by x, I will get

(3- (sqrt(9x^(2)+10))/x)/(3-(1000/x))

where do I go from here?

Debsta · #7 $Old$ November 6th, 2009, 10:22 PM

Make this bit: sqrt(9x^(2)+10))/x into sqrt(9x^(2)+10)/x^2) ie get your /x bit under the sqrt as well.

redsoxfan325 · #8 $Old$ November 7th, 2009, 01:44 AM

Quote:

Originally Posted by ctran $View Post$

How would you compute this limit?

lim (3x-sqrt(9x^(2)+10))/ (3x-1000)
x-> -inf

We actually have to be a little bit careful about dividing by $x$ with the $-\infty$ here. Look at this:

$\lim_{x\to-\infty}\frac{3x-\sqrt{9x^2+10}}{3x-1000}=\lim_{x\to-\infty}\frac{3x-\sqrt{9x^2(1+\frac{10}{9x^2})}}{3x(1-\frac{1000}{3x})}{\color{red}~=}$ $\lim_{x\to-\infty}\frac{3x-3x\sqrt{1+\frac{10}{9x^2}}}{3x(1-\frac{1000}{3x})}=\lim_{x\to-\infty}\frac{1-\sqrt{1+\frac{10}{9x^2}}}{1-\frac{1000}{3x}}=\frac{1-\sqrt{1}}{1}=0$

This limit here evaluates to $0$ . But $0$ is not the correct answer. The reason being is that when taking the $9x^2$ out from under the square root sign, it turns into $3x$ , which is affected by the negative sign in front of $\infty$ , as opposed to the $9x^2$ which is not. The best way to do this problem is to recognize that:

$\lim_{x\to-\infty}\frac{3x-\sqrt{9x^2+10}}{3x-1000}=\lim_{x\to\infty}\frac{3(-x)-\sqrt{9(-x)^2+10}}{3(-x)-1000}=\lim_{x\to\infty}\frac{-3x-\sqrt{9x^2(1+\frac{10}{9x^2})}}{-3x(1+\frac{1000}{3x})}$ $\lim_{x\to\infty}\frac{-3x\left(1+\sqrt{1+\frac{10}{9x^2}}\right)}{-3x(1+\frac{1000}{3x})}=\frac{1+\sqrt{1}}{1}=2$

which is the correct answer.

mr fantastic · #9 $Old$ November 7th, 2009, 04:22 AM

Quote:

Originally Posted by ctran $View Post$

How would you compute this limit?

lim (3x-sqrt(9x^(2)+10))/ (3x-1000)
x-> -inf

My advice is to first make the substitution t = -x so that it becomes a limit t --> +oo. Then divide top and bottom by t.