Mister Exam
Other calculators:
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How to use it?
- Limit of the function:
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Limit of (4+x^2)/(-6+2*x)
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Limit of ((1+x)/(1+2*x))^x
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Limit of (n/(1+n))^(5+3*n)
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Limit of (9^x-8^x)/asin(3*x)
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Identical expressions
- (n/(one +n))^(five + three *n)
- (n divide by (1 plus n)) to the power of (5 plus 3 multiply by n)
- (n divide by (one plus n)) to the power of (five plus three multiply by n)
- (n/(1+n))(5+3*n)
- n/1+n5+3*n
- (n/(1+n))^(5+3n)
- (n/(1+n))(5+3n)
- n/1+n5+3n
- n/1+n^5+3n
- (n divide by (1+n))^(5+3*n)
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Similar expressions
- (n/(1-n))^(5+3*n)
- (n/(1+n))^(5-3*n)
Limit of the function (n/(1+n))^(5+3*n)
The solution
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5 + 3*n
/ n \
lim |-----|
n->oo\1 + n/
$$\lim_{n \to \infty} \left(\frac{n}{n + 1}\right)^{3 n + 5}$$
Limit((n/(1 + n))^(5 + 3*n), n, oo, dir='-')
Detail solution
Let's take the limit
$$\lim_{n \to \infty} \left(\frac{n}{n + 1}\right)^{3 n + 5}$$
transform
$$\lim_{n \to \infty} \left(\frac{n}{n + 1}\right)^{3 n + 5}$$
=
$$\lim_{n \to \infty} \left(\frac{\left(n + 1\right) - 1}{n + 1}\right)^{3 n + 5}$$
=
$$\lim_{n \to \infty} \left(- \frac{1}{n + 1} + \frac{n + 1}{n + 1}\right)^{3 n + 5}$$
=
$$\lim_{n \to \infty} \left(1 - \frac{1}{n + 1}\right)^{3 n + 5}$$
=
do replacement
$$u = \frac{n + 1}{-1}$$
then
$$\lim_{n \to \infty} \left(1 - \frac{1}{n + 1}\right)^{3 n + 5}$$ =
=
$$\lim_{u \to \infty} \left(1 + \frac{1}{u}\right)^{2 - 3 u}$$
=
$$\lim_{u \to \infty}\left(\left(1 + \frac{1}{u}\right)^{2} \left(1 + \frac{1}{u}\right)^{- 3 u}\right)$$
=
$$\lim_{u \to \infty} \left(1 + \frac{1}{u}\right)^{2} \lim_{u \to \infty} \left(1 + \frac{1}{u}\right)^{- 3 u}$$
=
$$\lim_{u \to \infty} \left(1 + \frac{1}{u}\right)^{- 3 u}$$
=
$$\left(\left(\lim_{u \to \infty} \left(1 + \frac{1}{u}\right)^{u}\right)\right)^{-3}$$
The limit
$$\lim_{u \to \infty} \left(1 + \frac{1}{u}\right)^{u}$$
is second remarkable limit, is equal to e ~ 2.718281828459045
then
$$\left(\left(\lim_{u \to \infty} \left(1 + \frac{1}{u}\right)^{u}\right)\right)^{-3} = e^{-3}$$
The final answer:
$$\lim_{n \to \infty} \left(\frac{n}{n + 1}\right)^{3 n + 5} = e^{-3}$$
$$\lim_{n \to \infty} \left(\frac{n}{n + 1}\right)^{3 n + 5}$$
transform
$$\lim_{n \to \infty} \left(\frac{n}{n + 1}\right)^{3 n + 5}$$
=
$$\lim_{n \to \infty} \left(\frac{\left(n + 1\right) - 1}{n + 1}\right)^{3 n + 5}$$
=
$$\lim_{n \to \infty} \left(- \frac{1}{n + 1} + \frac{n + 1}{n + 1}\right)^{3 n + 5}$$
=
$$\lim_{n \to \infty} \left(1 - \frac{1}{n + 1}\right)^{3 n + 5}$$
=
do replacement
$$u = \frac{n + 1}{-1}$$
then
$$\lim_{n \to \infty} \left(1 - \frac{1}{n + 1}\right)^{3 n + 5}$$ =
=
$$\lim_{u \to \infty} \left(1 + \frac{1}{u}\right)^{2 - 3 u}$$
=
$$\lim_{u \to \infty}\left(\left(1 + \frac{1}{u}\right)^{2} \left(1 + \frac{1}{u}\right)^{- 3 u}\right)$$
=
$$\lim_{u \to \infty} \left(1 + \frac{1}{u}\right)^{2} \lim_{u \to \infty} \left(1 + \frac{1}{u}\right)^{- 3 u}$$
=
$$\lim_{u \to \infty} \left(1 + \frac{1}{u}\right)^{- 3 u}$$
=
$$\left(\left(\lim_{u \to \infty} \left(1 + \frac{1}{u}\right)^{u}\right)\right)^{-3}$$
The limit
$$\lim_{u \to \infty} \left(1 + \frac{1}{u}\right)^{u}$$
is second remarkable limit, is equal to e ~ 2.718281828459045
then
$$\left(\left(\lim_{u \to \infty} \left(1 + \frac{1}{u}\right)^{u}\right)\right)^{-3} = e^{-3}$$
The final answer:
$$\lim_{n \to \infty} \left(\frac{n}{n + 1}\right)^{3 n + 5} = e^{-3}$$
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
The graph