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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.
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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anonymous
 one year ago
Best ResponseYou've already chosen the best response.0dw:1441823645902:dw

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0hint: sqrt(x^2) = x , for x <0

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0be 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

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I 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.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0can you show your whole solution?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I got it. Thank you so much! cx

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0alright good job then lol

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0there is more than one way to solve this, here is one approach $$ \large \lim_{x \to \infty} \frac{3x + 1}{ \sqrt{x^2+1} } \\ \large \lim_{x \to \infty} \frac{3x + 1}{ \sqrt{x^2+1} } \cdot \frac{ \frac{1}{x} }{\frac{1}{\sqrt{x^2}}} $$

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0because 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
 one year ago
Best ResponseYou've already chosen the best response.0as an example \( \sqrt{(3)^2} = (3) = 3 \)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0$$\large \lim_{x \to \infty} \frac{3x + 1}{ \sqrt{x^2+1} } \\ \ \\ \large \lim_{x \to \infty} \frac{3x + 1}{ \sqrt{x^2+1} } \cdot \frac{ \frac{1}{x} }{\frac{1}{\sqrt{x^2}}} \\ \ \\ \Large \lim_{x \to \infty} \frac{ \frac{3x}{x} + \frac{1}{x} }{ \sqrt{\frac{x^2}{x^2}+\frac{1}{x^2}} } $$

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0thats 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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