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(6+n)/(4+n)

Limit of the function (6+n)/(4+n)

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     /6 + n\
 lim |-----|
n->oo\4 + n/
limn(n+6n+4)\lim_{n \to \infty}\left(\frac{n + 6}{n + 4}\right)
Limit((6 + n)/(4 + n), n, oo, dir='-')
Detail solution
Let's take the limit
limn(n+6n+4)\lim_{n \to \infty}\left(\frac{n + 6}{n + 4}\right)
Let's divide numerator and denominator by n:
limn(n+6n+4)\lim_{n \to \infty}\left(\frac{n + 6}{n + 4}\right) =
limn(1+6n1+4n)\lim_{n \to \infty}\left(\frac{1 + \frac{6}{n}}{1 + \frac{4}{n}}\right)
Do Replacement
u=1nu = \frac{1}{n}
then
limn(1+6n1+4n)=limu0+(6u+14u+1)\lim_{n \to \infty}\left(\frac{1 + \frac{6}{n}}{1 + \frac{4}{n}}\right) = \lim_{u \to 0^+}\left(\frac{6 u + 1}{4 u + 1}\right)
=
06+104+1=1\frac{0 \cdot 6 + 1}{0 \cdot 4 + 1} = 1

The final answer:
limn(n+6n+4)=1\lim_{n \to \infty}\left(\frac{n + 6}{n + 4}\right) = 1
Lopital's rule
We have indeterminateness of type
oo/oo,

i.e. limit for the numerator is
limn(n+6)=\lim_{n \to \infty}\left(n + 6\right) = \infty
and limit for the denominator is
limn(n+4)=\lim_{n \to \infty}\left(n + 4\right) = \infty
Let's take derivatives of the numerator and denominator until we eliminate indeterninateness.
limn(n+6n+4)\lim_{n \to \infty}\left(\frac{n + 6}{n + 4}\right)
=
limn(ddn(n+6)ddn(n+4))\lim_{n \to \infty}\left(\frac{\frac{d}{d n} \left(n + 6\right)}{\frac{d}{d n} \left(n + 4\right)}\right)
=
limn1\lim_{n \to \infty} 1
=
limn1\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
Rapid solution [src]
1
11
Other limits n→0, -oo, +oo, 1
limn(n+6n+4)=1\lim_{n \to \infty}\left(\frac{n + 6}{n + 4}\right) = 1
limn0(n+6n+4)=32\lim_{n \to 0^-}\left(\frac{n + 6}{n + 4}\right) = \frac{3}{2}
More at n→0 from the left
limn0+(n+6n+4)=32\lim_{n \to 0^+}\left(\frac{n + 6}{n + 4}\right) = \frac{3}{2}
More at n→0 from the right
limn1(n+6n+4)=75\lim_{n \to 1^-}\left(\frac{n + 6}{n + 4}\right) = \frac{7}{5}
More at n→1 from the left
limn1+(n+6n+4)=75\lim_{n \to 1^+}\left(\frac{n + 6}{n + 4}\right) = \frac{7}{5}
More at n→1 from the right
limn(n+6n+4)=1\lim_{n \to -\infty}\left(\frac{n + 6}{n + 4}\right) = 1
More at n→-oo
The graph
Limit of the function (6+n)/(4+n)