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(1+n)^4/n^4
Limit of the function/ (1+n)^4/n^4

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(1+n)^4/n^4

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(1 + n)^4/(n^4)
 
(1 + n)^((4/n)^4)
 
(1 + n)^((4/n)^4)

Limit of the function (1+n)^4/n^4

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The solution

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     /       4\
     |(1 + n) |
 lim |--------|
n->oo|    4   |
     \   n    /
lim⁡n→∞((n+1)4n4)\lim_{n \to \infty}\left(\frac{\left(n + 1\right)^{4}}{n^{4}}\right) 
Limit((1 + n)^4/(n^4), n, oo, dir='-')
Detail solution
Let's take the limit
lim⁡n→∞((n+1)4n4)\lim_{n \to \infty}\left(\frac{\left(n + 1\right)^{4}}{n^{4}}\right)
Let's divide numerator and denominator by n^4:
lim⁡n→∞((n+1)4n4)\lim_{n \to \infty}\left(\frac{\left(n + 1\right)^{4}}{n^{4}}\right) =
lim⁡n→∞(1+4n+6n2+4n3+1n41)\lim_{n \to \infty}\left(\frac{1 + \frac{4}{n} + \frac{6}{n^{2}} + \frac{4}{n^{3}} + \frac{1}{n^{4}}}{1}\right)
Do Replacement
u=1nu = \frac{1}{n}
then
lim⁡n→∞(1+4n+6n2+4n3+1n41)=lim⁡u→0+(u4+4u3+6u2+4u+1)\lim_{n \to \infty}\left(\frac{1 + \frac{4}{n} + \frac{6}{n^{2}} + \frac{4}{n^{3}} + \frac{1}{n^{4}}}{1}\right) = \lim_{u \to 0^+}\left(u^{4} + 4 u^{3} + 6 u^{2} + 4 u + 1\right)
=
04+4⋅0+4⋅03+6⋅02+1=10^{4} + 4 \cdot 0 + 4 \cdot 0^{3} + 6 \cdot 0^{2} + 1 = 1

The final answer:
lim⁡n→∞((n+1)4n4)=1\lim_{n \to \infty}\left(\frac{\left(n + 1\right)^{4}}{n^{4}}\right) = 1
Lopital's rule
We have indeterminateness of type
oo/oo,

i.e. limit for the numerator is
lim⁡n→∞(n+1)4=∞\lim_{n \to \infty} \left(n + 1\right)^{4} = \infty
and limit for the denominator is
lim⁡n→∞n4=∞\lim_{n \to \infty} n^{4} = \infty
Let's take derivatives of the numerator and denominator until we eliminate indeterninateness.
lim⁡n→∞((n+1)4n4)\lim_{n \to \infty}\left(\frac{\left(n + 1\right)^{4}}{n^{4}}\right)
=
Let's transform the function under the limit a few
lim⁡n→∞((n+1)4n4)\lim_{n \to \infty}\left(\frac{\left(n + 1\right)^{4}}{n^{4}}\right)
=
lim⁡n→∞(ddn(n+1)4ddnn4)\lim_{n \to \infty}\left(\frac{\frac{d}{d n} \left(n + 1\right)^{4}}{\frac{d}{d n} n^{4}}\right)
=
lim⁡n→∞((n+1)3n3)\lim_{n \to \infty}\left(\frac{\left(n + 1\right)^{3}}{n^{3}}\right)
=
lim⁡n→∞(ddn4(n+1)3ddn4n3)\lim_{n \to \infty}\left(\frac{\frac{d}{d n} 4 \left(n + 1\right)^{3}}{\frac{d}{d n} 4 n^{3}}\right)
=
lim⁡n→∞((n+1)2n2)\lim_{n \to \infty}\left(\frac{\left(n + 1\right)^{2}}{n^{2}}\right)
=
lim⁡n→∞(ddn12(n+1)2ddn12n2)\lim_{n \to \infty}\left(\frac{\frac{d}{d n} 12 \left(n + 1\right)^{2}}{\frac{d}{d n} 12 n^{2}}\right)
=
lim⁡n→∞(24n+2424n)\lim_{n \to \infty}\left(\frac{24 n + 24}{24 n}\right)
=
lim⁡n→∞(ddn(24n+24)ddn24n)\lim_{n \to \infty}\left(\frac{\frac{d}{d n} \left(24 n + 24\right)}{\frac{d}{d n} 24 n}\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) 4 time(s)
The graph
02468-8-6-4-2-1010020000
Plot the graph
Rapid solution [src] 
1
11
Expand and simplify
Other limits n→0, -oo, +oo, 1
lim⁡n→∞((n+1)4n4)=1\lim_{n \to \infty}\left(\frac{\left(n + 1\right)^{4}}{n^{4}}\right) = 1
lim⁡n→0−((n+1)4n4)=∞\lim_{n \to 0^-}\left(\frac{\left(n + 1\right)^{4}}{n^{4}}\right) = \infty
More at n→0 from the left
lim⁡n→0+((n+1)4n4)=∞\lim_{n \to 0^+}\left(\frac{\left(n + 1\right)^{4}}{n^{4}}\right) = \infty
More at n→0 from the right
lim⁡n→1−((n+1)4n4)=16\lim_{n \to 1^-}\left(\frac{\left(n + 1\right)^{4}}{n^{4}}\right) = 16
More at n→1 from the left
lim⁡n→1+((n+1)4n4)=16\lim_{n \to 1^+}\left(\frac{\left(n + 1\right)^{4}}{n^{4}}\right) = 16
More at n→1 from the right
lim⁡n→−∞((n+1)4n4)=1\lim_{n \to -\infty}\left(\frac{\left(n + 1\right)^{4}}{n^{4}}\right) = 1
More at n→-oo
The graph
Limit of the function (1+n)^4/n^4
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