Mister Exam
Other calculators:
-
How to use it?
- Limit of the function:
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Limit of (1+3*x)^(5/x)
-
Limit of e^(1+3*x)*(-1+x)
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Limit of x*sin(2*x)/3
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Limit of ((3+n)/(5+n))^(4+n)
- Graphing y =:
- ((3+n)/(5+n))^(4+n)
-
Identical expressions
- ((three +n)/(five +n))^(four +n)
- ((3 plus n) divide by (5 plus n)) to the power of (4 plus n)
- ((three plus n) divide by (five plus n)) to the power of (four plus n)
- ((3+n)/(5+n))(4+n)
- 3+n/5+n4+n
- 3+n/5+n^4+n
- ((3+n) divide by (5+n))^(4+n)
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Similar expressions
- ((3+n)/(5-n))^(4+n)
- ((3-n)/(5+n))^(4+n)
- ((3+n)/(5+n))^(4-n)
Limit of the function ((3+n)/(5+n))^(4+n)
The solution
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4 + n
/3 + n\
lim |-----|
n->oo\5 + n/
$$\lim_{n \to \infty} \left(\frac{n + 3}{n + 5}\right)^{n + 4}$$
Limit(((3 + n)/(5 + n))^(4 + n), n, oo, dir='-')
Detail solution
Let's take the limit
$$\lim_{n \to \infty} \left(\frac{n + 3}{n + 5}\right)^{n + 4}$$
transform
$$\lim_{n \to \infty} \left(\frac{n + 3}{n + 5}\right)^{n + 4}$$
=
$$\lim_{n \to \infty} \left(\frac{\left(n + 5\right) - 2}{n + 5}\right)^{n + 4}$$
=
$$\lim_{n \to \infty} \left(- \frac{2}{n + 5} + \frac{n + 5}{n + 5}\right)^{n + 4}$$
=
$$\lim_{n \to \infty} \left(1 - \frac{2}{n + 5}\right)^{n + 4}$$
=
do replacement
$$u = \frac{n + 5}{-2}$$
then
$$\lim_{n \to \infty} \left(1 - \frac{2}{n + 5}\right)^{n + 4}$$ =
=
$$\lim_{u \to \infty} \left(1 + \frac{1}{u}\right)^{- 2 u - 1}$$
=
$$\lim_{u \to \infty}\left(\frac{\left(1 + \frac{1}{u}\right)^{- 2 u}}{1 + \frac{1}{u}}\right)$$
=
$$\lim_{u \to \infty} \frac{1}{1 + \frac{1}{u}} \lim_{u \to \infty} \left(1 + \frac{1}{u}\right)^{- 2 u}$$
=
$$\lim_{u \to \infty} \left(1 + \frac{1}{u}\right)^{- 2 u}$$
=
$$\left(\left(\lim_{u \to \infty} \left(1 + \frac{1}{u}\right)^{u}\right)\right)^{-2}$$
The limit
$$\lim_{u \to \infty} \left(1 + \frac{1}{u}\right)^{u}$$
is second remarkable limit, is equal to e ~ 2.718281828459045
then
$$\left(\left(\lim_{u \to \infty} \left(1 + \frac{1}{u}\right)^{u}\right)\right)^{-2} = e^{-2}$$
The final answer:
$$\lim_{n \to \infty} \left(\frac{n + 3}{n + 5}\right)^{n + 4} = e^{-2}$$
$$\lim_{n \to \infty} \left(\frac{n + 3}{n + 5}\right)^{n + 4}$$
transform
$$\lim_{n \to \infty} \left(\frac{n + 3}{n + 5}\right)^{n + 4}$$
=
$$\lim_{n \to \infty} \left(\frac{\left(n + 5\right) - 2}{n + 5}\right)^{n + 4}$$
=
$$\lim_{n \to \infty} \left(- \frac{2}{n + 5} + \frac{n + 5}{n + 5}\right)^{n + 4}$$
=
$$\lim_{n \to \infty} \left(1 - \frac{2}{n + 5}\right)^{n + 4}$$
=
do replacement
$$u = \frac{n + 5}{-2}$$
then
$$\lim_{n \to \infty} \left(1 - \frac{2}{n + 5}\right)^{n + 4}$$ =
=
$$\lim_{u \to \infty} \left(1 + \frac{1}{u}\right)^{- 2 u - 1}$$
=
$$\lim_{u \to \infty}\left(\frac{\left(1 + \frac{1}{u}\right)^{- 2 u}}{1 + \frac{1}{u}}\right)$$
=
$$\lim_{u \to \infty} \frac{1}{1 + \frac{1}{u}} \lim_{u \to \infty} \left(1 + \frac{1}{u}\right)^{- 2 u}$$
=
$$\lim_{u \to \infty} \left(1 + \frac{1}{u}\right)^{- 2 u}$$
=
$$\left(\left(\lim_{u \to \infty} \left(1 + \frac{1}{u}\right)^{u}\right)\right)^{-2}$$
The limit
$$\lim_{u \to \infty} \left(1 + \frac{1}{u}\right)^{u}$$
is second remarkable limit, is equal to e ~ 2.718281828459045
then
$$\left(\left(\lim_{u \to \infty} \left(1 + \frac{1}{u}\right)^{u}\right)\right)^{-2} = e^{-2}$$
The final answer:
$$\lim_{n \to \infty} \left(\frac{n + 3}{n + 5}\right)^{n + 4} = e^{-2}$$
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
Other limits n→0, -oo, +oo, 1
$$\lim_{n \to \infty} \left(\frac{n + 3}{n + 5}\right)^{n + 4} = e^{-2}$$
$$\lim_{n \to 0^-} \left(\frac{n + 3}{n + 5}\right)^{n + 4} = \frac{81}{625}$$
More at n→0 from the left
$$\lim_{n \to 0^+} \left(\frac{n + 3}{n + 5}\right)^{n + 4} = \frac{81}{625}$$
More at n→0 from the right
$$\lim_{n \to 1^-} \left(\frac{n + 3}{n + 5}\right)^{n + 4} = \frac{32}{243}$$
More at n→1 from the left
$$\lim_{n \to 1^+} \left(\frac{n + 3}{n + 5}\right)^{n + 4} = \frac{32}{243}$$
More at n→1 from the right
$$\lim_{n \to -\infty} \left(\frac{n + 3}{n + 5}\right)^{n + 4} = e^{-2}$$
More at n→-oo
$$\lim_{n \to 0^-} \left(\frac{n + 3}{n + 5}\right)^{n + 4} = \frac{81}{625}$$
More at n→0 from the left
$$\lim_{n \to 0^+} \left(\frac{n + 3}{n + 5}\right)^{n + 4} = \frac{81}{625}$$
More at n→0 from the right
$$\lim_{n \to 1^-} \left(\frac{n + 3}{n + 5}\right)^{n + 4} = \frac{32}{243}$$
More at n→1 from the left
$$\lim_{n \to 1^+} \left(\frac{n + 3}{n + 5}\right)^{n + 4} = \frac{32}{243}$$
More at n→1 from the right
$$\lim_{n \to -\infty} \left(\frac{n + 3}{n + 5}\right)^{n + 4} = e^{-2}$$
More at n→-oo
The graph