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x^2-4*x
Limit of the function/ x^2-4*x

Limit of the function x^2-4*x

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     / 2      \
 lim \x  - 4*x/
x->oo
lim⁡x→∞(x2−4x)\lim_{x \to \infty}\left(x^{2} - 4 x\right) 
Limit(x^2 - 4*x, x, oo, dir='-')
Detail solution
Let's take the limit
lim⁡x→∞(x2−4x)\lim_{x \to \infty}\left(x^{2} - 4 x\right)
Let's divide numerator and denominator by x^2:
lim⁡x→∞(x2−4x)\lim_{x \to \infty}\left(x^{2} - 4 x\right) =
lim⁡x→∞(1−4x1x2)\lim_{x \to \infty}\left(\frac{1 - \frac{4}{x}}{\frac{1}{x^{2}}}\right)
Do Replacement
u=1xu = \frac{1}{x}
then
lim⁡x→∞(1−4x1x2)=lim⁡u→0+(1−4uu2)\lim_{x \to \infty}\left(\frac{1 - \frac{4}{x}}{\frac{1}{x^{2}}}\right) = \lim_{u \to 0^+}\left(\frac{1 - 4 u}{u^{2}}\right)
=
1−00=∞\frac{1 - 0}{0} = \infty

The final answer:
lim⁡x→∞(x2−4x)=∞\lim_{x \to \infty}\left(x^{2} - 4 x\right) = \infty
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
02468-8-6-4-2-1010-200200
Plot the graph
Other limits x→0, -oo, +oo, 1
lim⁡x→∞(x2−4x)=∞\lim_{x \to \infty}\left(x^{2} - 4 x\right) = \infty
lim⁡x→0−(x2−4x)=0\lim_{x \to 0^-}\left(x^{2} - 4 x\right) = 0
More at x→0 from the left
lim⁡x→0+(x2−4x)=0\lim_{x \to 0^+}\left(x^{2} - 4 x\right) = 0
More at x→0 from the right
lim⁡x→1−(x2−4x)=−3\lim_{x \to 1^-}\left(x^{2} - 4 x\right) = -3
More at x→1 from the left
lim⁡x→1+(x2−4x)=−3\lim_{x \to 1^+}\left(x^{2} - 4 x\right) = -3
More at x→1 from the right
lim⁡x→−∞(x2−4x)=∞\lim_{x \to -\infty}\left(x^{2} - 4 x\right) = \infty
More at x→-oo
Rapid solution [src] 
oo
∞\infty
Expand and simplify
The graph
Limit of the function x^2-4*x
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