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Math Help - Finding a limit algebraically

  1. #1
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    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!
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  2. #2
    No one in Particular VonNemo19's Avatar
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    Quote Originally Posted by JessicaWade View Post
    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
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  3. #3
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    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!
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  4. #4
    No one in Particular VonNemo19's Avatar
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    Quote Originally Posted by JessicaWade View Post
    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.
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  5. #5
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    wouldn't you get "undefined" if you did it that way?
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  6. #6
    No one in Particular VonNemo19's Avatar
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    Quote Originally Posted by JessicaWade View Post
    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?
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  7. #7
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    how did you get to x^2-3 ? Since the 1 is under the radical?
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  8. #8
    No one in Particular VonNemo19's Avatar
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    Quote Originally Posted by JessicaWade View Post
    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
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  9. #9
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    ok sorry, it just seems like the radical disappeared?

    anyway.. don't you still get a zero in the denominator??
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  10. #10
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    ok, nevermind i see it.. just still confused about the radical going away, but that's alright.. thank you VERY much!
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  11. #11
    No one in Particular VonNemo19's Avatar
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    Quote Originally Posted by JessicaWade View Post
    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 View Post
    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}
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  12. #12
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    The light bulb just came on &I totally get it now! Thank you sooooooooo much!
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