# Finding a limit algebraically

• September 13th 2009, 11:15 AM
Finding a limit algebraically

lim as x approaches square root of 3. ((square root x^2+1) - 2)/(x - square root 3)

hopefully that equation makes sense, I don't know how to make all the symbols &such..

• September 13th 2009, 11:19 AM
VonNemo19
Quote:

lim as x approaches square root of 3. ((square root x^2+1) - 2)/(x - square root 3)

hopefully that equation makes sense, I don't know how to make all the symbols &such..

Rationalize the denomiator by multiplying both Num. and Den. by the conjugate of the Num.

Ie Multiply by $(\sqrt{x^2+1}+2)$ and see where that takes you
• September 13th 2009, 11:26 AM
ok, i got (4 square root 3)/((square root 3) - 3)

that doesn't seem right.. did i do something wrong, or does it reduce more?? i tried taking the limit using my ti-89 &it said square root 3 over 2.. but i've got to show all my work.. thanks for your reply!
• September 13th 2009, 11:27 AM
VonNemo19
Quote:

ok, i got (4 square root 3)/((square root 3) - 3)

that doesn't seem right.. did i do something wrong, or does it reduce more?? i tried taking the limit using my ti-89 &it said square root 3 over 2.. but i've got to show all my work.. thanks for your reply!

I apologize. I've updated my last post. Sorry.
• September 13th 2009, 11:29 AM
wouldn't you get "undefined" if you did it that way?
• September 13th 2009, 11:44 AM
VonNemo19
Quote:

wouldn't you get "undefined" if you did it that way?

Im sorry, for some reason my brain isn't working right right now. Here's the solution.
1.Multiply by the conjugate of the Num and simplify. Then multiply by the conjugate of the Dem. and simplify.
$\lim_{x\to\sqrt{3}}\frac{\sqrt{x^2+1}-2}{x-\sqrt{3}}=\lim_{x\to\sqrt{3}}\frac{(\sqrt{x^2+1}-2)(\sqrt{x^2+1}+2)}{(x-\sqrt{3})(\sqrt{x^2+1}+2)}$

$=\lim_{x\to\sqrt{3}}\frac{(x^2-3)(x+\sqrt{3})}{(x-\sqrt{3})(\sqrt{x^2+1}+2)(x+\sqrt{3})}=\lim_{x\to\ sqrt{3}}\frac{(x^2-3)(x+\sqrt{3})}{(x^2-3)(\sqrt{x^2+1}+2)}$

Can you see?
• September 13th 2009, 11:49 AM
how did you get to x^2-3 ? Since the 1 is under the radical?
• September 13th 2009, 11:54 AM
VonNemo19
Quote:

how did you get to x^2-3 ? Since the 1 is under the radical?

Because

$[(\sqrt{x^2+1})-2][(\sqrt{x^2+1})+2]=$ $(\sqrt{x^2+1})^2-2\sqrt{x^2+1}+2\sqrt{x^2+1}-4=x^2+1-4=x^2-3$
• September 13th 2009, 12:02 PM
ok sorry, it just seems like the radical disappeared?

anyway.. don't you still get a zero in the denominator??
• September 13th 2009, 12:07 PM
ok, nevermind i see it.. just still confused about the radical going away, but that's alright.. thank you VERY much!
• September 13th 2009, 12:13 PM
VonNemo19
Quote:

ok sorry, it just seems like the radical disappeared?

The radical dissapears because $(\sqrt{x^2+1})^2=x^2+1$. Inverse processes, ya know?

Quote:

No because the $(x^2-3)'s$ cancel leaving
$\lim_{x\to\sqrt{3}}\frac{x+\sqrt{3}}{\sqrt{x^2+1}+ 2}=\frac{\sqrt{3}+\sqrt{3}}{\sqrt{\sqrt{3}^2+2}+1}$ $=\frac{2\sqrt{3}}{\sqrt{3+1}+2}=\frac{2\sqrt{3}}{4 }=\frac{\sqrt{3}}{2}$