Mister Exam
Other calculators:
-
How to use it?
- Limit of the function:
-
Limit of ((-2+3*x)/(1+3*x))^(2*x)
-
Limit of (1-cos(4*x))/(1-cos(8*x))
-
Limit of (3+x^2-4*x)/(-9+x^2)
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Limit of (1+x^2-4*x)/(1+2*x)
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Identical expressions
- (one +n)^ four /n^ four
- (1 plus n) to the power of 4 divide by n to the power of 4
- (one plus n) to the power of four divide by n to the power of four
- (1+n)4/n4
- 1+n4/n4
- (1+n)⁴/n⁴
- 1+n^4/n^4
- (1+n)^4 divide by n^4
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Similar expressions
- (1-n)^4/n^4
-
What you mean?
- (1 + n)^4/(n^4)
- (1 + n)^((4/n)^4)
- (1 + n)^((4/n)^4)
You entered:
(1+n)^4/n^4
What you mean?
Limit of the function (1+n)^4/n^4
The solution
You have entered
[src]
/ 4\
|(1 + n) |
lim |--------|
n->oo| 4 |
\ n /
$$\lim_{n \to \infty}\left(\frac{\left(n + 1\right)^{4}}{n^{4}}\right)$$
Limit((1 + n)^4/(n^4), n, oo, dir='-')
Detail solution
Let's take the limit
$$\lim_{n \to \infty}\left(\frac{\left(n + 1\right)^{4}}{n^{4}}\right)$$
Let's divide numerator and denominator by n^4:
$$\lim_{n \to \infty}\left(\frac{\left(n + 1\right)^{4}}{n^{4}}\right)$$ =
$$\lim_{n \to \infty}\left(\frac{1 + \frac{4}{n} + \frac{6}{n^{2}} + \frac{4}{n^{3}} + \frac{1}{n^{4}}}{1}\right)$$
Do Replacement
$$u = \frac{1}{n}$$
then
$$\lim_{n \to \infty}\left(\frac{1 + \frac{4}{n} + \frac{6}{n^{2}} + \frac{4}{n^{3}} + \frac{1}{n^{4}}}{1}\right) = \lim_{u \to 0^+}\left(u^{4} + 4 u^{3} + 6 u^{2} + 4 u + 1\right)$$
=
$$0^{4} + 4 \cdot 0 + 4 \cdot 0^{3} + 6 \cdot 0^{2} + 1 = 1$$
The final answer:
$$\lim_{n \to \infty}\left(\frac{\left(n + 1\right)^{4}}{n^{4}}\right) = 1$$
$$\lim_{n \to \infty}\left(\frac{\left(n + 1\right)^{4}}{n^{4}}\right)$$
Let's divide numerator and denominator by n^4:
$$\lim_{n \to \infty}\left(\frac{\left(n + 1\right)^{4}}{n^{4}}\right)$$ =
$$\lim_{n \to \infty}\left(\frac{1 + \frac{4}{n} + \frac{6}{n^{2}} + \frac{4}{n^{3}} + \frac{1}{n^{4}}}{1}\right)$$
Do Replacement
$$u = \frac{1}{n}$$
then
$$\lim_{n \to \infty}\left(\frac{1 + \frac{4}{n} + \frac{6}{n^{2}} + \frac{4}{n^{3}} + \frac{1}{n^{4}}}{1}\right) = \lim_{u \to 0^+}\left(u^{4} + 4 u^{3} + 6 u^{2} + 4 u + 1\right)$$
=
$$0^{4} + 4 \cdot 0 + 4 \cdot 0^{3} + 6 \cdot 0^{2} + 1 = 1$$
The final answer:
$$\lim_{n \to \infty}\left(\frac{\left(n + 1\right)^{4}}{n^{4}}\right) = 1$$
Lopital's rule
We have indeterminateness of type
i.e. limit for the numerator is
$$\lim_{n \to \infty} \left(n + 1\right)^{4} = \infty$$
and limit for the denominator is
$$\lim_{n \to \infty} n^{4} = \infty$$
Let's take derivatives of the numerator and denominator until we eliminate indeterninateness.
$$\lim_{n \to \infty}\left(\frac{\left(n + 1\right)^{4}}{n^{4}}\right)$$
=
Let's transform the function under the limit a few
$$\lim_{n \to \infty}\left(\frac{\left(n + 1\right)^{4}}{n^{4}}\right)$$
=
$$\lim_{n \to \infty}\left(\frac{\frac{d}{d n} \left(n + 1\right)^{4}}{\frac{d}{d n} n^{4}}\right)$$
=
$$\lim_{n \to \infty}\left(\frac{\left(n + 1\right)^{3}}{n^{3}}\right)$$
=
$$\lim_{n \to \infty}\left(\frac{\frac{d}{d n} 4 \left(n + 1\right)^{3}}{\frac{d}{d n} 4 n^{3}}\right)$$
=
$$\lim_{n \to \infty}\left(\frac{\left(n + 1\right)^{2}}{n^{2}}\right)$$
=
$$\lim_{n \to \infty}\left(\frac{\frac{d}{d n} 12 \left(n + 1\right)^{2}}{\frac{d}{d n} 12 n^{2}}\right)$$
=
$$\lim_{n \to \infty}\left(\frac{24 n + 24}{24 n}\right)$$
=
$$\lim_{n \to \infty}\left(\frac{\frac{d}{d n} \left(24 n + 24\right)}{\frac{d}{d n} 24 n}\right)$$
=
$$\lim_{n \to \infty} 1$$
=
$$\lim_{n \to \infty} 1$$
=
$$1$$
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)
oo/oo,
i.e. limit for the numerator is
$$\lim_{n \to \infty} \left(n + 1\right)^{4} = \infty$$
and limit for the denominator is
$$\lim_{n \to \infty} n^{4} = \infty$$
Let's take derivatives of the numerator and denominator until we eliminate indeterninateness.
$$\lim_{n \to \infty}\left(\frac{\left(n + 1\right)^{4}}{n^{4}}\right)$$
=
Let's transform the function under the limit a few
$$\lim_{n \to \infty}\left(\frac{\left(n + 1\right)^{4}}{n^{4}}\right)$$
=
$$\lim_{n \to \infty}\left(\frac{\frac{d}{d n} \left(n + 1\right)^{4}}{\frac{d}{d n} n^{4}}\right)$$
=
$$\lim_{n \to \infty}\left(\frac{\left(n + 1\right)^{3}}{n^{3}}\right)$$
=
$$\lim_{n \to \infty}\left(\frac{\frac{d}{d n} 4 \left(n + 1\right)^{3}}{\frac{d}{d n} 4 n^{3}}\right)$$
=
$$\lim_{n \to \infty}\left(\frac{\left(n + 1\right)^{2}}{n^{2}}\right)$$
=
$$\lim_{n \to \infty}\left(\frac{\frac{d}{d n} 12 \left(n + 1\right)^{2}}{\frac{d}{d n} 12 n^{2}}\right)$$
=
$$\lim_{n \to \infty}\left(\frac{24 n + 24}{24 n}\right)$$
=
$$\lim_{n \to \infty}\left(\frac{\frac{d}{d n} \left(24 n + 24\right)}{\frac{d}{d n} 24 n}\right)$$
=
$$\lim_{n \to \infty} 1$$
=
$$\lim_{n \to \infty} 1$$
=
$$1$$
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
Other limits n→0, -oo, +oo, 1
$$\lim_{n \to \infty}\left(\frac{\left(n + 1\right)^{4}}{n^{4}}\right) = 1$$
$$\lim_{n \to 0^-}\left(\frac{\left(n + 1\right)^{4}}{n^{4}}\right) = \infty$$
More at n→0 from the left
$$\lim_{n \to 0^+}\left(\frac{\left(n + 1\right)^{4}}{n^{4}}\right) = \infty$$
More at n→0 from the right
$$\lim_{n \to 1^-}\left(\frac{\left(n + 1\right)^{4}}{n^{4}}\right) = 16$$
More at n→1 from the left
$$\lim_{n \to 1^+}\left(\frac{\left(n + 1\right)^{4}}{n^{4}}\right) = 16$$
More at n→1 from the right
$$\lim_{n \to -\infty}\left(\frac{\left(n + 1\right)^{4}}{n^{4}}\right) = 1$$
More at n→-oo
$$\lim_{n \to 0^-}\left(\frac{\left(n + 1\right)^{4}}{n^{4}}\right) = \infty$$
More at n→0 from the left
$$\lim_{n \to 0^+}\left(\frac{\left(n + 1\right)^{4}}{n^{4}}\right) = \infty$$
More at n→0 from the right
$$\lim_{n \to 1^-}\left(\frac{\left(n + 1\right)^{4}}{n^{4}}\right) = 16$$
More at n→1 from the left
$$\lim_{n \to 1^+}\left(\frac{\left(n + 1\right)^{4}}{n^{4}}\right) = 16$$
More at n→1 from the right
$$\lim_{n \to -\infty}\left(\frac{\left(n + 1\right)^{4}}{n^{4}}\right) = 1$$
More at n→-oo
The graph