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x/(3+x^2)

Limit of the function x/(3+x^2)

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     /  x   \
 lim |------|
x->oo|     2|
     \3 + x /
limx(xx2+3)\lim_{x \to \infty}\left(\frac{x}{x^{2} + 3}\right)
Limit(x/(3 + x^2), x, oo, dir='-')
Detail solution
Let's take the limit
limx(xx2+3)\lim_{x \to \infty}\left(\frac{x}{x^{2} + 3}\right)
Let's divide numerator and denominator by x^2:
limx(xx2+3)\lim_{x \to \infty}\left(\frac{x}{x^{2} + 3}\right) =
limx(1x(1+3x2))\lim_{x \to \infty}\left(\frac{1}{x \left(1 + \frac{3}{x^{2}}\right)}\right)
Do Replacement
u=1xu = \frac{1}{x}
then
limx(1x(1+3x2))=limu0+(u3u2+1)\lim_{x \to \infty}\left(\frac{1}{x \left(1 + \frac{3}{x^{2}}\right)}\right) = \lim_{u \to 0^+}\left(\frac{u}{3 u^{2} + 1}\right)
=
0302+1=0\frac{0}{3 \cdot 0^{2} + 1} = 0

The final answer:
limx(xx2+3)=0\lim_{x \to \infty}\left(\frac{x}{x^{2} + 3}\right) = 0
Lopital's rule
We have indeterminateness of type
oo/oo,

i.e. limit for the numerator is
limxx=\lim_{x \to \infty} x = \infty
and limit for the denominator is
limx(x2+3)=\lim_{x \to \infty}\left(x^{2} + 3\right) = \infty
Let's take derivatives of the numerator and denominator until we eliminate indeterninateness.
limx(xx2+3)\lim_{x \to \infty}\left(\frac{x}{x^{2} + 3}\right)
=
limx(ddxxddx(x2+3))\lim_{x \to \infty}\left(\frac{\frac{d}{d x} x}{\frac{d}{d x} \left(x^{2} + 3\right)}\right)
=
limx(12x)\lim_{x \to \infty}\left(\frac{1}{2 x}\right)
=
limx(12x)\lim_{x \to \infty}\left(\frac{1}{2 x}\right)
=
00
It can be seen that we have applied Lopital's rule (we have taken derivatives with respect to the numerator and denominator) 1 time(s)
The graph
02468-8-6-4-2-10100.5-0.5
Rapid solution [src]
0
00
Other limits x→0, -oo, +oo, 1
limx(xx2+3)=0\lim_{x \to \infty}\left(\frac{x}{x^{2} + 3}\right) = 0
limx0(xx2+3)=0\lim_{x \to 0^-}\left(\frac{x}{x^{2} + 3}\right) = 0
More at x→0 from the left
limx0+(xx2+3)=0\lim_{x \to 0^+}\left(\frac{x}{x^{2} + 3}\right) = 0
More at x→0 from the right
limx1(xx2+3)=14\lim_{x \to 1^-}\left(\frac{x}{x^{2} + 3}\right) = \frac{1}{4}
More at x→1 from the left
limx1+(xx2+3)=14\lim_{x \to 1^+}\left(\frac{x}{x^{2} + 3}\right) = \frac{1}{4}
More at x→1 from the right
limx(xx2+3)=0\lim_{x \to -\infty}\left(\frac{x}{x^{2} + 3}\right) = 0
More at x→-oo
The graph
Limit of the function x/(3+x^2)