Finding a limit algebraically

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September 13th 2009, 11:15 AM
JessicaWade

Finding a limit algebraically

OK, I can't figure out what to do since there is a radical in both the numerator and denominator...please help!

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..

thanks in advance!
September 13th 2009, 11:19 AM
VonNemo19

Quote:

Originally Posted by JessicaWade

OK, I can't figure out what to do since there is a radical in both the numerator and denominator...please help!

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..

thanks in advance!

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
JessicaWade

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:

Originally Posted by JessicaWade

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
JessicaWade

wouldn't you get "undefined" if you did it that way?
September 13th 2009, 11:44 AM
VonNemo19

Quote:

Originally Posted by JessicaWade

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
JessicaWade

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

Quote:

Originally Posted by JessicaWade

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
JessicaWade

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
JessicaWade

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:

Originally Posted by JessicaWade

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:

Originally Posted by JessicaWade

anyway.. don't you still get a zero in the denominator??

No because the $(x^2-3)'s$ cancel leaving

$\lim_{x\to\sqrt{3}}\frac{x+\sqrt{3}}{\sqrt{x^2+1}+ 2}=\frac{\sqrt{3}+\sqrt{3}}{\sqrt{\sqrt{3}^2+2}+1}$ $=\frac{2\sqrt{3}}{\sqrt{3+1}+2}=\frac{2\sqrt{3}}{4 }=\frac{\sqrt{3}}{2}$
September 13th 2009, 12:17 PM
JessicaWade

The light bulb just came on &I totally get it now! Thank you sooooooooo much!