Ap calculus: lim x-> -(negative) infinity (x+?x^2+3x)

[joy08]

I'm having trouble with finding the limit for this problem.
Find the limit : lim x-> -(negative) infinity (x+?x^2+3x)

 

[Deleted member 4993]

Re: Ap calculus

I'm having trouble with finding the limit for this problem.
Find the limit : lim x-> -(negative) infinity (x+?x^2+3x)

This is a tricky one....

Hint:

The answer is -3/2

First multiply and divide by (x-?x^2+3x)

Then let x = -m and let m -> +infinity

 

[o_O]

Re: Ap calculus

Mmm I don't think substitution is necessary.

\(\displaystyle \lim_{x \to -\infty} \left( \left(x + \sqrt{x^{2} + 3x}\right) \cdot \frac{x - \sqrt{x^{2} + 3x}}{x - \sqrt{x^{2} + 3x}} \right)\)

\(\displaystyle \lim_{x \to -\infty} \left(\frac{x^{2} - x^{2} - 3x}{x - \sqrt{x^{2} + 3x}} \right)\)

\(\displaystyle \lim_{x \to -\infty} \left( \frac{-3x}{x - \sqrt{x^{2} + 3x}} \right)\)

Then multiply top and bottom by \(\displaystyle \frac{1}{x}\) keeping in mind that x is negative so: \(\displaystyle x = - |x| = -\sqrt{x^{2}}\)