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Limit of the function
:
Limit of (-1+e^(2*x))/(3*x)
Limit of (20-17*x+3*x^2)/(36-25*x+4*x^2)
Limit of (3+n)/(1+n)
Limit of 1+13*x/5
Sum of series
:
1/(1+n^2)
Identical expressions
one /(one +n^ two)
1 divide by (1 plus n squared )
one divide by (one plus n to the power of two)
1/(1+n2)
1/1+n2
1/(1+n²)
1/(1+n to the power of 2)
1/1+n^2
1 divide by (1+n^2)
Similar expressions
1/(1-n^2)
Limit of the function
/
1/(1+n^2)
Limit of the function 1/(1+n^2)
at
→
Calculate the limit!
v
For end points:
---------
From the left (x0-)
From the right (x0+)
The graph:
from
to
Piecewise:
{
enter the piecewise function here
The solution
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[src]
1 lim ------ n->oo 2 1 + n
$$\lim_{n \to \infty} \frac{1}{n^{2} + 1}$$
Limit(1/(1 + n^2), n, oo, dir='-')
Detail solution
Let's take the limit
$$\lim_{n \to \infty} \frac{1}{n^{2} + 1}$$
Let's divide numerator and denominator by n^2:
$$\lim_{n \to \infty} \frac{1}{n^{2} + 1}$$ =
$$\lim_{n \to \infty}\left(\frac{1}{n^{2} \left(1 + \frac{1}{n^{2}}\right)}\right)$$
Do Replacement
$$u = \frac{1}{n}$$
then
$$\lim_{n \to \infty}\left(\frac{1}{n^{2} \left(1 + \frac{1}{n^{2}}\right)}\right) = \lim_{u \to 0^+}\left(\frac{u^{2}}{u^{2} + 1}\right)$$
=
$$\frac{0^{2}}{0^{2} + 1} = 0$$
The final answer:
$$\lim_{n \to \infty} \frac{1}{n^{2} + 1} = 0$$
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
Plot the graph
Rapid solution
[src]
0
$$0$$
Expand and simplify
Other limits n→0, -oo, +oo, 1
$$\lim_{n \to \infty} \frac{1}{n^{2} + 1} = 0$$
$$\lim_{n \to 0^-} \frac{1}{n^{2} + 1} = 1$$
More at n→0 from the left
$$\lim_{n \to 0^+} \frac{1}{n^{2} + 1} = 1$$
More at n→0 from the right
$$\lim_{n \to 1^-} \frac{1}{n^{2} + 1} = \frac{1}{2}$$
More at n→1 from the left
$$\lim_{n \to 1^+} \frac{1}{n^{2} + 1} = \frac{1}{2}$$
More at n→1 from the right
$$\lim_{n \to -\infty} \frac{1}{n^{2} + 1} = 0$$
More at n→-oo
The graph