Limits help (1 Viewer)

findsome1 · May 3, 2010

Aye I dont get the whole deal of negative infinite limits. Could you please explain this example:

Find lim x ---> negative infinite x^2+x+3/x^2

The answer is

1 from below... o.0

thongetsu · May 3, 2010

I think you divide by the highest power and the x and 3 become zero (2unit limits) since its infinity. And since it's negative infinity the graph comes from below.

ohexploitable · May 3, 2010

Think of the number line, if you go to the right you approach infinity, if you go to the left you approach negative infinity.

1 from below means that it's approaching 1 from the left side, so the limit is -5, -4, -3, -2, -1 , 0, 0.9, 0.99, 0.999 etc.

omgmynameislong · May 4, 2010

Long divide.

fullonoob · May 4, 2010

$f(x)=\frac {x^2 + x + 3}{x^2}\\ =\lim_{x\to -\infty}\frac {x^2 + x + 3}{x^2}\\ =\lim_{x\to -\infty}x + \frac{1}{x} + \frac {3}{x^2}\\ =1^- \ \ \ since\ others\ are\ limited\ to\ -0.00000000001\\ hence\ the\ value\ is\ 0.9999999999999$

farr explotable your gif is laggy as hell lol

give me rating on this comp please, go engineering forum, then go on the thread makeitso.

cutemouse · May 4, 2010

fullonoob said:
$f(x)=\frac {x^2 + x + 3}{x^2}\\ =\lim_{x\to -\infty}\frac {x^2 + x + 3}{x^2}\\ =\lim_{x\to -\infty}x + \frac{1}{x} + \frac {3}{x^2}\\ =1^- \ \ \ since\ others\ are\ limited\ to\ -0.00000000001\\ hence\ the\ value\ is\ 0.9999999999999$

I think you mean 1 instead of 'x'.

thongetsu · May 4, 2010

fullonoob said:
farr explotable your gif is laggy as hell lol.

noob net.

Limits help (1 Viewer)

findsome1

Member

thongetsu

Where aren't I?

ohexploitable

hey

omgmynameislong

Member

fullonoob

fail engrish? unpossible!

cutemouse

Account Closed

thongetsu

Where aren't I?

Users Who Are Viewing This Thread (Users: 0, Guests: 1)