# limits at negative infinity

• Oct 22nd 2012, 06:36 PM
kingsolomonsgrave
limits at negative infinity
Attachment 25349

when taking the limit as x goes to negative infinity, and we divided by negative sqrt(x^2) why does the term 5x not become -5?

the negative is associated only with the second term in the denominator but not the 5, why?

also why is the denominator not divided by a negative?
• Oct 23rd 2012, 12:15 AM
Siron
Re: limits at negative infinity
This is what happens:
$\lim_{x \to \infty} \frac{5x+\sqrt{x^2+2x}}{4+x} = \lim_{x \to \infty} \frac{5x+\sqrt{x^2\left(1+\frac{2}{x}\right)}}{x \left(\frac{4}{x}+1\right)}$
$= \lim_{x \to \infty} \frac{5x+|x|\sqrt{1+\frac{2}{x}}}{x\left(\frac{4}{ x}+1\right)}$

Suppose $x \to +\infty$ then $|x| = x$ thus
$\lim_{x \to +\infty} \frac{x\left(5+\sqrt{1+\frac{2}{x}}\right)}{x\left (\frac{4}{x}+1\right)} = \lim_{x \to +\infty} \frac{5+\sqrt{1+\frac{2}{x}}}{\frac{4}{x}+1} = 6$

Suppose $x \to -\infty$ then $|x|=-x$
Now (check it) the limit will be equal to 4
• Oct 23rd 2012, 08:42 AM
FernandoRevilla
Re: limits at negative infinity
Perphas this also can help you: substituting for example $x=-3$ in $\frac{5x+\sqrt{x^2+2x}}{4+x}$ we get $\frac{5(-3)+\sqrt{(-3)^2+2(-3)}}{4+(-3)}$. Now, divide numerator by $-\sqrt{(-3)^2}\;(=-3)$ and denonimator by $-3$ . What do you obtain? (substtitute now $-3$ by $x$)