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4/(1+x^2)
Limit of the function/ 4/(1+x^2)

Limit of the function 4/(1+x^2)

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     /  4   \
 lim |------|
x->oo|     2|
     \1 + x /
lim⁡x→∞(4x2+1)\lim_{x \to \infty}\left(\frac{4}{x^{2} + 1}\right) 
Limit(4/(1 + x^2), x, oo, dir='-')
Detail solution
Let's take the limit
lim⁡x→∞(4x2+1)\lim_{x \to \infty}\left(\frac{4}{x^{2} + 1}\right)
Let's divide numerator and denominator by x^2:
lim⁡x→∞(4x2+1)\lim_{x \to \infty}\left(\frac{4}{x^{2} + 1}\right) =
lim⁡x→∞(41x21+1x2)\lim_{x \to \infty}\left(\frac{4 \frac{1}{x^{2}}}{1 + \frac{1}{x^{2}}}\right)
Do Replacement
u=1xu = \frac{1}{x}
then
lim⁡x→∞(41x21+1x2)=lim⁡u→0+(4u2u2+1)\lim_{x \to \infty}\left(\frac{4 \frac{1}{x^{2}}}{1 + \frac{1}{x^{2}}}\right) = \lim_{u \to 0^+}\left(\frac{4 u^{2}}{u^{2} + 1}\right)
=
4⋅0202+1=0\frac{4 \cdot 0^{2}}{0^{2} + 1} = 0

The final answer:
lim⁡x→∞(4x2+1)=0\lim_{x \to \infty}\left(\frac{4}{x^{2} + 1}\right) = 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
02468-8-6-4-2-101005
Plot the graph
Other limits x→0, -oo, +oo, 1
lim⁡x→∞(4x2+1)=0\lim_{x \to \infty}\left(\frac{4}{x^{2} + 1}\right) = 0
lim⁡x→0−(4x2+1)=4\lim_{x \to 0^-}\left(\frac{4}{x^{2} + 1}\right) = 4
More at x→0 from the left
lim⁡x→0+(4x2+1)=4\lim_{x \to 0^+}\left(\frac{4}{x^{2} + 1}\right) = 4
More at x→0 from the right
lim⁡x→1−(4x2+1)=2\lim_{x \to 1^-}\left(\frac{4}{x^{2} + 1}\right) = 2
More at x→1 from the left
lim⁡x→1+(4x2+1)=2\lim_{x \to 1^+}\left(\frac{4}{x^{2} + 1}\right) = 2
More at x→1 from the right
lim⁡x→−∞(4x2+1)=0\lim_{x \to -\infty}\left(\frac{4}{x^{2} + 1}\right) = 0
More at x→-oo
Rapid solution [src] 
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The graph
Limit of the function 4/(1+x^2)
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