## 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.

1. anonymous

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2. anonymous

hint: sqrt(x^2) = -x , for x <0

3. anonymous

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

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. anonymous

can you show your whole solution?

6. anonymous

I got it. Thank you so much! cx

7. anonymous

alright good job then lol

8. anonymous

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

9. anonymous

that's interesting

10. 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

11. anonymous

as an example $$\sqrt{(-3)^2} = -(-3) = 3$$

12. anonymous

$$\large \lim_{x \to -\infty} \frac{3x + 1}{ \sqrt{x^2+1} } \\ \ \\ \large \lim_{x \to -\infty} \frac{3x + 1}{ \sqrt{x^2+1} } \cdot \frac{ \frac{1}{x} }{\frac{1}{-\sqrt{x^2}}} \\ \ \\ \Large \lim_{x \to -\infty} \frac{ \frac{3x}{x} + \frac{1}{x} }{ -\sqrt{\frac{x^2}{x^2}+\frac{1}{x^2}} }$$

13. 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