anonymous
  • anonymous
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.
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
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anonymous
  • anonymous
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anonymous
  • anonymous
hint: sqrt(x^2) = -x , for x <0
hwyl
  • hwyl
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

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More answers

anonymous
  • anonymous
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.
hwyl
  • hwyl
can you show your whole solution?
anonymous
  • anonymous
I got it. Thank you so much! cx
hwyl
  • hwyl
alright good job then lol
anonymous
  • anonymous
there is more than one way to solve this, here is one approach $$ \large \lim_{x \to -\infty} \frac{3x + 1}{ \sqrt{x^2+1} } \\ \large \lim_{x \to -\infty} \frac{3x + 1}{ \sqrt{x^2+1} } \cdot \frac{ \frac{1}{x} }{\frac{1}{-\sqrt{x^2}}} $$
hwyl
  • hwyl
that's interesting
anonymous
  • anonymous
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
anonymous
  • anonymous
as an example \( \sqrt{(-3)^2} = -(-3) = 3 \)
anonymous
  • anonymous
$$\large \lim_{x \to -\infty} \frac{3x + 1}{ \sqrt{x^2+1} } \\ \ \\ \large \lim_{x \to -\infty} \frac{3x + 1}{ \sqrt{x^2+1} } \cdot \frac{ \frac{1}{x} }{\frac{1}{-\sqrt{x^2}}} \\ \ \\ \Large \lim_{x \to -\infty} \frac{ \frac{3x}{x} + \frac{1}{x} }{ -\sqrt{\frac{x^2}{x^2}+\frac{1}{x^2}} } $$
anonymous
  • anonymous
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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