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anonymous

  • one year ago

How do you find the limit of 3x+1 / the square root of x^2 + x as x approaches negative infinity? I got 3, but the answer is -3.

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  1. anonymous
    • one year ago
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    |dw:1441823645902:dw|

  2. anonymous
    • one year ago
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    hint: sqrt(x^2) = -x , for x <0

  3. hwyl
    • one year ago
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    be aware that limits are from the negative and positive directions and in the problem you were given, it is approaching that which is less than zero

  4. anonymous
    • one year ago
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    I crossed out the + 1 in the numerator because it becomes irrelevant. Then squared the numerator and denominator. Leaving 3x^2 / x^2 + x. Then factored out an x leaving 3x / x. giving me 3.

  5. hwyl
    • one year ago
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    can you show your whole solution?

  6. anonymous
    • one year ago
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    I got it. Thank you so much! cx

  7. hwyl
    • one year ago
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    alright good job then lol

  8. anonymous
    • one year ago
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    there is more than one way to solve this, here is one approach $$ \large \lim_{x \to -\infty} \frac{3x + 1}{ \sqrt{x^2+1} } \\ \large \lim_{x \to -\infty} \frac{3x + 1}{ \sqrt{x^2+1} } \cdot \frac{ \frac{1}{x} }{\frac{1}{-\sqrt{x^2}}} $$

  9. hwyl
    • one year ago
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    that's interesting

  10. anonymous
    • one year ago
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    because for x<0 \( \sqrt{x^2} = -x \) so to get x, we have to multiply by -1 \( x= -\sqrt{x^2} \) for x<0

  11. anonymous
    • one year ago
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    as an example \( \sqrt{(-3)^2} = -(-3) = 3 \)

  12. anonymous
    • one year ago
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    $$\large \lim_{x \to -\infty} \frac{3x + 1}{ \sqrt{x^2+1} } \\ \ \\ \large \lim_{x \to -\infty} \frac{3x + 1}{ \sqrt{x^2+1} } \cdot \frac{ \frac{1}{x} }{\frac{1}{-\sqrt{x^2}}} \\ \ \\ \Large \lim_{x \to -\infty} \frac{ \frac{3x}{x} + \frac{1}{x} }{ -\sqrt{\frac{x^2}{x^2}+\frac{1}{x^2}} } $$

  13. anonymous
    • one year ago
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    thats the gist of it. then simplify you will get 3 + 1/x in the numerator, and -sqrt(1) in the denominator comes out to -3

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