y=(-x)/(x^2+1)
as x-infinity,
y=????
why

Question

y=(-x)/(x^2+1)
as x->infinity,
y=????
why

anonymous · Answer

Since $x^2$ dominates $x$ the limit as $x$ approaches infinity is 0

anonymous · Answer

$$-\frac{x}{x^2+1}= -\frac{1}{\displaystyle\frac{x^2}{x}+\frac{1}{x}} = -\frac{1}{x+\displaystyle\frac{1}{x}}$$$$\lim_{x \to \infty}-\frac{1}{x+\displaystyle\frac{1}{x}} = 0$$

anonymous · Answer

why divide x by x square

anonymous · Answer

It is to demonstrate that the denominator dominates the numerator as x approaches infinity.

anonymous · Answer

huh...??
is it we must always make the top a constant?

anonymous · Answer

What happens to $\displaystyle\frac{1}{x}$ for big $x$

anonymous · Answer

1/0

anonymous · Answer

i mean zero

anonymous · Answer

As x gets very large 1/x gets very small. As you can see above, 1/x on the bottom will disappear, but you will still have -1/x which will approach 0 as x gets large.

anonymous · Answer

so u look at things part by part? cant look at it straight?