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1/(x-x^2)

Limit of the function 1/(x-x^2)

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       1   
 lim ------
x->oo     2
     x - x 
limx1x2+x\lim_{x \to \infty} \frac{1}{- x^{2} + x}
Limit(1/(x - x^2), x, oo, dir='-')
Detail solution
Let's take the limit
limx1x2+x\lim_{x \to \infty} \frac{1}{- x^{2} + x}
Let's divide numerator and denominator by x^2:
limx1x2+x\lim_{x \to \infty} \frac{1}{- x^{2} + x} =
limx(1x2(1+1x))\lim_{x \to \infty}\left(\frac{1}{x^{2} \left(-1 + \frac{1}{x}\right)}\right)
Do Replacement
u=1xu = \frac{1}{x}
then
limx(1x2(1+1x))=limu0+(u2u1)\lim_{x \to \infty}\left(\frac{1}{x^{2} \left(-1 + \frac{1}{x}\right)}\right) = \lim_{u \to 0^+}\left(\frac{u^{2}}{u - 1}\right)
=
021=0\frac{0^{2}}{-1} = 0

The final answer:
limx1x2+x=0\lim_{x \to \infty} \frac{1}{- x^{2} + x} = 0
Lopital's rule
There is no sense to apply Lopital's rule to this function since there is no indeterminateness of 0/0 or oo/oo type
The graph
02468-8-6-4-2-1010-2525
Rapid solution [src]
0
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Other limits x→0, -oo, +oo, 1
limx1x2+x=0\lim_{x \to \infty} \frac{1}{- x^{2} + x} = 0
limx01x2+x=\lim_{x \to 0^-} \frac{1}{- x^{2} + x} = -\infty
More at x→0 from the left
limx0+1x2+x=\lim_{x \to 0^+} \frac{1}{- x^{2} + x} = \infty
More at x→0 from the right
limx11x2+x=\lim_{x \to 1^-} \frac{1}{- x^{2} + x} = \infty
More at x→1 from the left
limx1+1x2+x=\lim_{x \to 1^+} \frac{1}{- x^{2} + x} = -\infty
More at x→1 from the right
limx1x2+x=0\lim_{x \to -\infty} \frac{1}{- x^{2} + x} = 0
More at x→-oo
The graph
Limit of the function 1/(x-x^2)