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)/(3-x^2)
-
Limit of (-2+x^2-x)/(-2+x+3*x^2)
-
Limit of 5-9*x+3*x^2/2
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Limit of (-16+x^2)/(-64+x^3)
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Identical expressions
- ((one +x)/(two +x))^(one +x)
- ((1 plus x) divide by (2 plus x)) to the power of (1 plus x)
- ((one plus x) divide by (two plus x)) to the power of (one plus x)
- ((1+x)/(2+x))(1+x)
- 1+x/2+x1+x
- 1+x/2+x^1+x
- ((1+x) divide by (2+x))^(1+x)
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Similar expressions
- ((1+x)/(2-x))^(1+x)
- ((1-x)/(2+x))^(1+x)
- ((1+x)/(2+x))^(1-x)
Limit of the function ((1+x)/(2+x))^(1+x)
The solution
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1 + x
/1 + x\
lim |-----|
x->oo\2 + x/
$$\lim_{x \to \infty} \left(\frac{x + 1}{x + 2}\right)^{x + 1}$$
Limit(((1 + x)/(2 + x))^(1 + x), x, oo, dir='-')
Detail solution
Let's take the limit
$$\lim_{x \to \infty} \left(\frac{x + 1}{x + 2}\right)^{x + 1}$$
transform
$$\lim_{x \to \infty} \left(\frac{x + 1}{x + 2}\right)^{x + 1}$$
=
$$\lim_{x \to \infty} \left(\frac{\left(x + 2\right) - 1}{x + 2}\right)^{x + 1}$$
=
$$\lim_{x \to \infty} \left(- \frac{1}{x + 2} + \frac{x + 2}{x + 2}\right)^{x + 1}$$
=
$$\lim_{x \to \infty} \left(1 - \frac{1}{x + 2}\right)^{x + 1}$$
=
do replacement
$$u = \frac{x + 2}{-1}$$
then
$$\lim_{x \to \infty} \left(1 - \frac{1}{x + 2}\right)^{x + 1}$$ =
=
$$\lim_{u \to \infty} \left(1 + \frac{1}{u}\right)^{- u - 1}$$
=
$$\lim_{u \to \infty}\left(\frac{\left(1 + \frac{1}{u}\right)^{- u}}{1 + \frac{1}{u}}\right)$$
=
$$\lim_{u \to \infty} \frac{1}{1 + \frac{1}{u}} \lim_{u \to \infty} \left(1 + \frac{1}{u}\right)^{- u}$$
=
$$\lim_{u \to \infty} \left(1 + \frac{1}{u}\right)^{- u}$$
=
$$\left(\left(\lim_{u \to \infty} \left(1 + \frac{1}{u}\right)^{u}\right)\right)^{-1}$$
The limit
$$\lim_{u \to \infty} \left(1 + \frac{1}{u}\right)^{u}$$
is second remarkable limit, is equal to e ~ 2.718281828459045
then
$$\left(\left(\lim_{u \to \infty} \left(1 + \frac{1}{u}\right)^{u}\right)\right)^{-1} = e^{-1}$$
The final answer:
$$\lim_{x \to \infty} \left(\frac{x + 1}{x + 2}\right)^{x + 1} = e^{-1}$$
$$\lim_{x \to \infty} \left(\frac{x + 1}{x + 2}\right)^{x + 1}$$
transform
$$\lim_{x \to \infty} \left(\frac{x + 1}{x + 2}\right)^{x + 1}$$
=
$$\lim_{x \to \infty} \left(\frac{\left(x + 2\right) - 1}{x + 2}\right)^{x + 1}$$
=
$$\lim_{x \to \infty} \left(- \frac{1}{x + 2} + \frac{x + 2}{x + 2}\right)^{x + 1}$$
=
$$\lim_{x \to \infty} \left(1 - \frac{1}{x + 2}\right)^{x + 1}$$
=
do replacement
$$u = \frac{x + 2}{-1}$$
then
$$\lim_{x \to \infty} \left(1 - \frac{1}{x + 2}\right)^{x + 1}$$ =
=
$$\lim_{u \to \infty} \left(1 + \frac{1}{u}\right)^{- u - 1}$$
=
$$\lim_{u \to \infty}\left(\frac{\left(1 + \frac{1}{u}\right)^{- u}}{1 + \frac{1}{u}}\right)$$
=
$$\lim_{u \to \infty} \frac{1}{1 + \frac{1}{u}} \lim_{u \to \infty} \left(1 + \frac{1}{u}\right)^{- u}$$
=
$$\lim_{u \to \infty} \left(1 + \frac{1}{u}\right)^{- u}$$
=
$$\left(\left(\lim_{u \to \infty} \left(1 + \frac{1}{u}\right)^{u}\right)\right)^{-1}$$
The limit
$$\lim_{u \to \infty} \left(1 + \frac{1}{u}\right)^{u}$$
is second remarkable limit, is equal to e ~ 2.718281828459045
then
$$\left(\left(\lim_{u \to \infty} \left(1 + \frac{1}{u}\right)^{u}\right)\right)^{-1} = e^{-1}$$
The final answer:
$$\lim_{x \to \infty} \left(\frac{x + 1}{x + 2}\right)^{x + 1} = e^{-1}$$
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