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Math Help - Why am I getting this L'Hospital limit wrong?!

  1. #1
    s3a
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    Why am I getting this L'Hospital limit wrong?!

    According to Wolfram Alpha (the website), the answer is 3 however, when I do it without rationalizing the numerator with steps I deem valid, I get the wrong answer! I'm just hoping someone could either spot my mistake or tell me why my method doesn't work if that is the case.

    Any input would be greatly appreciated!
    Thanks in advance!
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  2. #2
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    \lim_{x \to \infty} (\sqrt{x^2+6x+5} - x) \cdot \frac{\sqrt{x^2+6x+5} + x}{\sqrt{x^2+6x+5} + x}

    \lim_{x \to \infty} \frac{x^2+6x+5 - x^2}{\sqrt{x^2+6x+5} + x}

    \lim_{x \to \infty} \frac{6x+5}{\sqrt{x^2+6x+5} + x}

    \lim_{x \to \infty} \frac{6+\frac{5}{x}}{\sqrt{1+\frac{6}{x}+\frac{5}{  x^2}} + 1} = 3
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  3. #3
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    First change to: \sqrt{x^2+6x+25}-x=\frac{6x+25}{\sqrt{x^2+6x+25}+x}
    Now divide by x.

    BTW: Go back to Wolfram Alpha.
    When you get the result. Click on the 'Show Steps' button int the upper right hand.
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  4. #4
    s3a
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    Yes, I know about the steps on Wolfram Alpha but I wanted to know why my method did not work.
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  5. #5
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    Quote Originally Posted by s3a View Post
    Yes, I know about the steps on Wolfram Alpha but I wanted to know why my method did not work.
    because you forgot the chain rule.
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