Mister Exam
Other calculators:
-
How to use it?
- Limit of the function:
-
Limit of (1+3*x)^(5/x)
-
Limit of x^2/(-2+sqrt(4+x^2))
-
Limit of ((5+x^2-6*x)/(5+x^2-5*x))^(2+3*x)
-
Limit of (2+x^2+3*x)/(-4+x^2)
- Integral of d{x}:
-
e^x/(1+e^x)
- Graphing y =:
- e^x/(1+e^x)
- Derivative of:
-
e^x/(1+e^x)
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Identical expressions
- e^x/(one +e^x)
- e to the power of x divide by (1 plus e to the power of x)
- e to the power of x divide by (one plus e to the power of x)
- ex/(1+ex)
- ex/1+ex
- e^x/1+e^x
- e^x divide by (1+e^x)
-
Similar expressions
- e^x/(1-e^x)
- (-1+3*e^x)/(1+e^x)
Limit of the function e^x/(1+e^x)
The solution
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[src]
/ x \
| E |
lim |------|
x->oo| x|
\1 + E /
$$\lim_{x \to \infty}\left(\frac{e^{x}}{e^{x} + 1}\right)$$
Limit(E^x/(1 + E^x), x, oo, dir='-')
Lopital's rule
We have indeterminateness of type
i.e. limit for the numerator is
$$\lim_{x \to \infty} e^{x} = \infty$$
and limit for the denominator is
$$\lim_{x \to \infty}\left(e^{x} + 1\right) = \infty$$
Let's take derivatives of the numerator and denominator until we eliminate indeterninateness.
$$\lim_{x \to \infty}\left(\frac{e^{x}}{e^{x} + 1}\right)$$
=
Let's transform the function under the limit a few
$$\lim_{x \to \infty}\left(\frac{e^{x}}{e^{x} + 1}\right)$$
=
$$\lim_{x \to \infty}\left(\frac{\frac{d}{d x} e^{x}}{\frac{d}{d x} \left(e^{x} + 1\right)}\right)$$
=
$$\lim_{x \to \infty} 1$$
=
$$\lim_{x \to \infty} 1$$
=
$$1$$
It can be seen that we have applied Lopital's rule (we have taken derivatives with respect to the numerator and denominator) 1 time(s)
oo/oo,
i.e. limit for the numerator is
$$\lim_{x \to \infty} e^{x} = \infty$$
and limit for the denominator is
$$\lim_{x \to \infty}\left(e^{x} + 1\right) = \infty$$
Let's take derivatives of the numerator and denominator until we eliminate indeterninateness.
$$\lim_{x \to \infty}\left(\frac{e^{x}}{e^{x} + 1}\right)$$
=
Let's transform the function under the limit a few
$$\lim_{x \to \infty}\left(\frac{e^{x}}{e^{x} + 1}\right)$$
=
$$\lim_{x \to \infty}\left(\frac{\frac{d}{d x} e^{x}}{\frac{d}{d x} \left(e^{x} + 1\right)}\right)$$
=
$$\lim_{x \to \infty} 1$$
=
$$\lim_{x \to \infty} 1$$
=
$$1$$
It can be seen that we have applied Lopital's rule (we have taken derivatives with respect to the numerator and denominator) 1 time(s)
The graph
Other limits x→0, -oo, +oo, 1
$$\lim_{x \to \infty}\left(\frac{e^{x}}{e^{x} + 1}\right) = 1$$
$$\lim_{x \to 0^-}\left(\frac{e^{x}}{e^{x} + 1}\right) = \frac{1}{2}$$
More at x→0 from the left
$$\lim_{x \to 0^+}\left(\frac{e^{x}}{e^{x} + 1}\right) = \frac{1}{2}$$
More at x→0 from the right
$$\lim_{x \to 1^-}\left(\frac{e^{x}}{e^{x} + 1}\right) = \frac{e}{1 + e}$$
More at x→1 from the left
$$\lim_{x \to 1^+}\left(\frac{e^{x}}{e^{x} + 1}\right) = \frac{e}{1 + e}$$
More at x→1 from the right
$$\lim_{x \to -\infty}\left(\frac{e^{x}}{e^{x} + 1}\right) = 0$$
More at x→-oo
$$\lim_{x \to 0^-}\left(\frac{e^{x}}{e^{x} + 1}\right) = \frac{1}{2}$$
More at x→0 from the left
$$\lim_{x \to 0^+}\left(\frac{e^{x}}{e^{x} + 1}\right) = \frac{1}{2}$$
More at x→0 from the right
$$\lim_{x \to 1^-}\left(\frac{e^{x}}{e^{x} + 1}\right) = \frac{e}{1 + e}$$
More at x→1 from the left
$$\lim_{x \to 1^+}\left(\frac{e^{x}}{e^{x} + 1}\right) = \frac{e}{1 + e}$$
More at x→1 from the right
$$\lim_{x \to -\infty}\left(\frac{e^{x}}{e^{x} + 1}\right) = 0$$
More at x→-oo
The graph