Mister Exam
Other calculators:
-
How to use it?
- Limit of the function:
-
Limit of sin(x)/x+tan(x)
-
Limit of ((1+x)/x)^(1+x)
-
Limit of ((1+x)/(6+x))^(5-5*x)
-
Limit of e^(1/(-1+x))*(-1+x)
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Identical expressions
- ((one +x)/x)^(one +x)
- ((1 plus x) divide by x) to the power of (1 plus x)
- ((one plus x) divide by x) to the power of (one plus x)
- ((1+x)/x)(1+x)
- 1+x/x1+x
- 1+x/x^1+x
- ((1+x) divide by x)^(1+x)
-
Similar expressions
- ((1+x)/x)^(1-x)
- ((1-x)/x)^(1+x)
Limit of the function ((1+x)/x)^(1+x)
The solution
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[src]
1 + x
/1 + x\
lim |-----|
x->oo\ x /
$$\lim_{x \to \infty} \left(\frac{x + 1}{x}\right)^{x + 1}$$
Limit(((1 + x)/x)^(1 + x), x, oo, dir='-')
Detail solution
Let's take the limit
$$\lim_{x \to \infty} \left(\frac{x + 1}{x}\right)^{x + 1}$$
transform
$$\lim_{x \to \infty} \left(\frac{x + 1}{x}\right)^{x + 1}$$
=
$$\lim_{x \to \infty} \left(\frac{x + 1}{x}\right)^{x + 1}$$
=
$$\lim_{x \to \infty} \left(\frac{x}{x} + \frac{1}{x}\right)^{x + 1}$$
=
$$\lim_{x \to \infty} \left(1 + \frac{1}{x}\right)^{x + 1}$$
=
do replacement
$$u = \frac{x}{1}$$
then
$$\lim_{x \to \infty} \left(1 + \frac{1}{x}\right)^{x + 1}$$ =
=
$$\lim_{u \to \infty} \left(1 + \frac{1}{u}\right)^{u + 1}$$
=
$$\lim_{u \to \infty}\left(\left(1 + \frac{1}{u}\right)^{1} \left(1 + \frac{1}{u}\right)^{u}\right)$$
=
$$\lim_{u \to \infty}\left(1 + \frac{1}{u}\right) \lim_{u \to \infty} \left(1 + \frac{1}{u}\right)^{u}$$
=
$$\lim_{u \to \infty} \left(1 + \frac{1}{u}\right)^{u}$$
=
$$\left(\left(\lim_{u \to \infty} \left(1 + \frac{1}{u}\right)^{u}\right)\right)$$
The limit
$$\lim_{u \to \infty} \left(1 + \frac{1}{u}\right)^{u}$$
is second remarkable limit, is equal to e ~ 2.718281828459045
then
$$\left(\left(\lim_{u \to \infty} \left(1 + \frac{1}{u}\right)^{u}\right)\right) = e$$
The final answer:
$$\lim_{x \to \infty} \left(\frac{x + 1}{x}\right)^{x + 1} = e$$
$$\lim_{x \to \infty} \left(\frac{x + 1}{x}\right)^{x + 1}$$
transform
$$\lim_{x \to \infty} \left(\frac{x + 1}{x}\right)^{x + 1}$$
=
$$\lim_{x \to \infty} \left(\frac{x + 1}{x}\right)^{x + 1}$$
=
$$\lim_{x \to \infty} \left(\frac{x}{x} + \frac{1}{x}\right)^{x + 1}$$
=
$$\lim_{x \to \infty} \left(1 + \frac{1}{x}\right)^{x + 1}$$
=
do replacement
$$u = \frac{x}{1}$$
then
$$\lim_{x \to \infty} \left(1 + \frac{1}{x}\right)^{x + 1}$$ =
=
$$\lim_{u \to \infty} \left(1 + \frac{1}{u}\right)^{u + 1}$$
=
$$\lim_{u \to \infty}\left(\left(1 + \frac{1}{u}\right)^{1} \left(1 + \frac{1}{u}\right)^{u}\right)$$
=
$$\lim_{u \to \infty}\left(1 + \frac{1}{u}\right) \lim_{u \to \infty} \left(1 + \frac{1}{u}\right)^{u}$$
=
$$\lim_{u \to \infty} \left(1 + \frac{1}{u}\right)^{u}$$
=
$$\left(\left(\lim_{u \to \infty} \left(1 + \frac{1}{u}\right)^{u}\right)\right)$$
The limit
$$\lim_{u \to \infty} \left(1 + \frac{1}{u}\right)^{u}$$
is second remarkable limit, is equal to e ~ 2.718281828459045
then
$$\left(\left(\lim_{u \to \infty} \left(1 + \frac{1}{u}\right)^{u}\right)\right) = e$$
The final answer:
$$\lim_{x \to \infty} \left(\frac{x + 1}{x}\right)^{x + 1} = e$$
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 x→0, -oo, +oo, 1
$$\lim_{x \to \infty} \left(\frac{x + 1}{x}\right)^{x + 1} = e$$
$$\lim_{x \to 0^-} \left(\frac{x + 1}{x}\right)^{x + 1} = -\infty$$
More at x→0 from the left
$$\lim_{x \to 0^+} \left(\frac{x + 1}{x}\right)^{x + 1} = \infty$$
More at x→0 from the right
$$\lim_{x \to 1^-} \left(\frac{x + 1}{x}\right)^{x + 1} = 4$$
More at x→1 from the left
$$\lim_{x \to 1^+} \left(\frac{x + 1}{x}\right)^{x + 1} = 4$$
More at x→1 from the right
$$\lim_{x \to -\infty} \left(\frac{x + 1}{x}\right)^{x + 1} = e$$
More at x→-oo
$$\lim_{x \to 0^-} \left(\frac{x + 1}{x}\right)^{x + 1} = -\infty$$
More at x→0 from the left
$$\lim_{x \to 0^+} \left(\frac{x + 1}{x}\right)^{x + 1} = \infty$$
More at x→0 from the right
$$\lim_{x \to 1^-} \left(\frac{x + 1}{x}\right)^{x + 1} = 4$$
More at x→1 from the left
$$\lim_{x \to 1^+} \left(\frac{x + 1}{x}\right)^{x + 1} = 4$$
More at x→1 from the right
$$\lim_{x \to -\infty} \left(\frac{x + 1}{x}\right)^{x + 1} = e$$
More at x→-oo
The graph