Mister Exam
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How to use it?
- Limit of the function:
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Limit of (1+1/x)^(2+5*x)
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Limit of (1+x^(-3))^(x^2)
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Limit of (x/(5+x))^(8+x)
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Limit of (x+x^3-x^2)/(1+x^2+x^4-x-x^3)
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Identical expressions
- (x/(five +x))^(eight +x)
- (x divide by (5 plus x)) to the power of (8 plus x)
- (x divide by (five plus x)) to the power of (eight plus x)
- (x/(5+x))(8+x)
- x/5+x8+x
- x/5+x^8+x
- (x divide by (5+x))^(8+x)
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Similar expressions
- (x/(5+x))^(8-x)
- (x/(5-x))^(8+x)
Limit of the function (x/(5+x))^(8+x)
The solution
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8 + x
/ x \
lim |-----|
x->oo\5 + x/
$$\lim_{x \to \infty} \left(\frac{x}{x + 5}\right)^{x + 8}$$
Limit((x/(5 + x))^(8 + x), x, oo, dir='-')
Detail solution
Let's take the limit
$$\lim_{x \to \infty} \left(\frac{x}{x + 5}\right)^{x + 8}$$
transform
$$\lim_{x \to \infty} \left(\frac{x}{x + 5}\right)^{x + 8}$$
=
$$\lim_{x \to \infty} \left(\frac{\left(x + 5\right) - 5}{x + 5}\right)^{x + 8}$$
=
$$\lim_{x \to \infty} \left(- \frac{5}{x + 5} + \frac{x + 5}{x + 5}\right)^{x + 8}$$
=
$$\lim_{x \to \infty} \left(1 - \frac{5}{x + 5}\right)^{x + 8}$$
=
do replacement
$$u = \frac{x + 5}{-5}$$
then
$$\lim_{x \to \infty} \left(1 - \frac{5}{x + 5}\right)^{x + 8}$$ =
=
$$\lim_{u \to \infty} \left(1 + \frac{1}{u}\right)^{3 - 5 u}$$
=
$$\lim_{u \to \infty}\left(\left(1 + \frac{1}{u}\right)^{3} \left(1 + \frac{1}{u}\right)^{- 5 u}\right)$$
=
$$\lim_{u \to \infty} \left(1 + \frac{1}{u}\right)^{3} \lim_{u \to \infty} \left(1 + \frac{1}{u}\right)^{- 5 u}$$
=
$$\lim_{u \to \infty} \left(1 + \frac{1}{u}\right)^{- 5 u}$$
=
$$\left(\left(\lim_{u \to \infty} \left(1 + \frac{1}{u}\right)^{u}\right)\right)^{-5}$$
The limit
$$\lim_{u \to \infty} \left(1 + \frac{1}{u}\right)^{u}$$
is second remarkable limit, is equal to e ~ 2.718281828459045
then
$$\left(\left(\lim_{u \to \infty} \left(1 + \frac{1}{u}\right)^{u}\right)\right)^{-5} = e^{-5}$$
The final answer:
$$\lim_{x \to \infty} \left(\frac{x}{x + 5}\right)^{x + 8} = e^{-5}$$
$$\lim_{x \to \infty} \left(\frac{x}{x + 5}\right)^{x + 8}$$
transform
$$\lim_{x \to \infty} \left(\frac{x}{x + 5}\right)^{x + 8}$$
=
$$\lim_{x \to \infty} \left(\frac{\left(x + 5\right) - 5}{x + 5}\right)^{x + 8}$$
=
$$\lim_{x \to \infty} \left(- \frac{5}{x + 5} + \frac{x + 5}{x + 5}\right)^{x + 8}$$
=
$$\lim_{x \to \infty} \left(1 - \frac{5}{x + 5}\right)^{x + 8}$$
=
do replacement
$$u = \frac{x + 5}{-5}$$
then
$$\lim_{x \to \infty} \left(1 - \frac{5}{x + 5}\right)^{x + 8}$$ =
=
$$\lim_{u \to \infty} \left(1 + \frac{1}{u}\right)^{3 - 5 u}$$
=
$$\lim_{u \to \infty}\left(\left(1 + \frac{1}{u}\right)^{3} \left(1 + \frac{1}{u}\right)^{- 5 u}\right)$$
=
$$\lim_{u \to \infty} \left(1 + \frac{1}{u}\right)^{3} \lim_{u \to \infty} \left(1 + \frac{1}{u}\right)^{- 5 u}$$
=
$$\lim_{u \to \infty} \left(1 + \frac{1}{u}\right)^{- 5 u}$$
=
$$\left(\left(\lim_{u \to \infty} \left(1 + \frac{1}{u}\right)^{u}\right)\right)^{-5}$$
The limit
$$\lim_{u \to \infty} \left(1 + \frac{1}{u}\right)^{u}$$
is second remarkable limit, is equal to e ~ 2.718281828459045
then
$$\left(\left(\lim_{u \to \infty} \left(1 + \frac{1}{u}\right)^{u}\right)\right)^{-5} = e^{-5}$$
The final answer:
$$\lim_{x \to \infty} \left(\frac{x}{x + 5}\right)^{x + 8} = e^{-5}$$
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