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(3+n)/(1+n)
Limit of the function/ (3+n)/(1+n)

Limit of the function (3+n)/(1+n)

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     /3 + n\
 lim |-----|
n->oo\1 + n/
lim⁡n→∞(n+3n+1)\lim_{n \to \infty}\left(\frac{n + 3}{n + 1}\right) 
Limit((3 + n)/(1 + n), n, oo, dir='-')
Detail solution
Let's take the limit
lim⁡n→∞(n+3n+1)\lim_{n \to \infty}\left(\frac{n + 3}{n + 1}\right)
Let's divide numerator and denominator by n:
lim⁡n→∞(n+3n+1)\lim_{n \to \infty}\left(\frac{n + 3}{n + 1}\right) =
lim⁡n→∞(1+3n1+1n)\lim_{n \to \infty}\left(\frac{1 + \frac{3}{n}}{1 + \frac{1}{n}}\right)
Do Replacement
u=1nu = \frac{1}{n}
then
lim⁡n→∞(1+3n1+1n)=lim⁡u→0+(3u+1u+1)\lim_{n \to \infty}\left(\frac{1 + \frac{3}{n}}{1 + \frac{1}{n}}\right) = \lim_{u \to 0^+}\left(\frac{3 u + 1}{u + 1}\right)
=
0⋅3+11=1\frac{0 \cdot 3 + 1}{1} = 1

The final answer:
lim⁡n→∞(n+3n+1)=1\lim_{n \to \infty}\left(\frac{n + 3}{n + 1}\right) = 1
Lopital's rule
We have indeterminateness of type
oo/oo,

i.e. limit for the numerator is
lim⁡n→∞(n+3)=∞\lim_{n \to \infty}\left(n + 3\right) = \infty
and limit for the denominator is
lim⁡n→∞(n+1)=∞\lim_{n \to \infty}\left(n + 1\right) = \infty
Let's take derivatives of the numerator and denominator until we eliminate indeterninateness.
lim⁡n→∞(n+3n+1)\lim_{n \to \infty}\left(\frac{n + 3}{n + 1}\right)
=
lim⁡n→∞(ddn(n+3)ddn(n+1))\lim_{n \to \infty}\left(\frac{\frac{d}{d n} \left(n + 3\right)}{\frac{d}{d n} \left(n + 1\right)}\right)
=
lim⁡n→∞1\lim_{n \to \infty} 1
=
lim⁡n→∞1\lim_{n \to \infty} 1
=
11
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
02468-8-6-4-2-1010-5050
Plot the graph
Other limits n→0, -oo, +oo, 1
lim⁡n→∞(n+3n+1)=1\lim_{n \to \infty}\left(\frac{n + 3}{n + 1}\right) = 1
lim⁡n→0−(n+3n+1)=3\lim_{n \to 0^-}\left(\frac{n + 3}{n + 1}\right) = 3
More at n→0 from the left
lim⁡n→0+(n+3n+1)=3\lim_{n \to 0^+}\left(\frac{n + 3}{n + 1}\right) = 3
More at n→0 from the right
lim⁡n→1−(n+3n+1)=2\lim_{n \to 1^-}\left(\frac{n + 3}{n + 1}\right) = 2
More at n→1 from the left
lim⁡n→1+(n+3n+1)=2\lim_{n \to 1^+}\left(\frac{n + 3}{n + 1}\right) = 2
More at n→1 from the right
lim⁡n→−∞(n+3n+1)=1\lim_{n \to -\infty}\left(\frac{n + 3}{n + 1}\right) = 1
More at n→-oo
Rapid solution [src] 
1
11
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The graph
Limit of the function (3+n)/(1+n)
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