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1+y^2
Limit of the function/ 1+y^2

Limit of the function 1+y^2

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     /     2\
 lim \1 + y /
y->oo
lim⁡y→∞(y2+1)\lim_{y \to \infty}\left(y^{2} + 1\right) 
Limit(1 + y^2, y, oo, dir='-')
Detail solution
Let's take the limit
lim⁡y→∞(y2+1)\lim_{y \to \infty}\left(y^{2} + 1\right)
Let's divide numerator and denominator by y^2:
lim⁡y→∞(y2+1)\lim_{y \to \infty}\left(y^{2} + 1\right) =
lim⁡y→∞(1+1y21y2)\lim_{y \to \infty}\left(\frac{1 + \frac{1}{y^{2}}}{\frac{1}{y^{2}}}\right)
Do Replacement
u=1yu = \frac{1}{y}
then
lim⁡y→∞(1+1y21y2)=lim⁡u→0+(u2+1u2)\lim_{y \to \infty}\left(\frac{1 + \frac{1}{y^{2}}}{\frac{1}{y^{2}}}\right) = \lim_{u \to 0^+}\left(\frac{u^{2} + 1}{u^{2}}\right)
=
02+10=∞\frac{0^{2} + 1}{0} = \infty

The final answer:
lim⁡y→∞(y2+1)=∞\lim_{y \to \infty}\left(y^{2} + 1\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
02468-8-6-4-2-10100200
Plot the graph
Rapid solution [src] 
oo
∞\infty
Expand and simplify
Other limits y→0, -oo, +oo, 1
lim⁡y→∞(y2+1)=∞\lim_{y \to \infty}\left(y^{2} + 1\right) = \infty
lim⁡y→0−(y2+1)=1\lim_{y \to 0^-}\left(y^{2} + 1\right) = 1
More at y→0 from the left
lim⁡y→0+(y2+1)=1\lim_{y \to 0^+}\left(y^{2} + 1\right) = 1
More at y→0 from the right
lim⁡y→1−(y2+1)=2\lim_{y \to 1^-}\left(y^{2} + 1\right) = 2
More at y→1 from the left
lim⁡y→1+(y2+1)=2\lim_{y \to 1^+}\left(y^{2} + 1\right) = 2
More at y→1 from the right
lim⁡y→−∞(y2+1)=∞\lim_{y \to -\infty}\left(y^{2} + 1\right) = \infty
More at y→-oo
The graph
Limit of the function 1+y^2
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