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.

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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?

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