Mister Exam
Other calculators:
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How to use it?
- Limit of the function:
-
Limit of x^(4/x)
-
Limit of (-6+x^2-x)/(-4+x^2)
-
Limit of (-6+x^2-x)/(-21+x+2*x^2)
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Limit of e^(3-x)*(-2+x)
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Identical expressions
- log(factorial(x))/x
- logarithm of (factorial(x)) divide by x
- logfactorialx/x
- log(factorial(x)) divide by x
-
Similar expressions
- log(factorial(x))/(x*log(x))
- log(x)*log(factorial(x))/(x*log(2)*log(5))
Limit of the function log(factorial(x))/x
The solution
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/log(x!)\ lim |-------| x->oo\ x /
$$\lim_{x \to \infty}\left(\frac{\log{\left(x! \right)}}{x}\right)$$
Limit(log(factorial(x))/x, x, oo, dir='-')
Lopital's rule
We have indeterminateness of type
i.e. limit for the numerator is
$$\lim_{x \to \infty} \log{\left(x! \right)} = \infty$$
and limit for the denominator is
$$\lim_{x \to \infty} x = \infty$$
Let's take derivatives of the numerator and denominator until we eliminate indeterninateness.
$$\lim_{x \to \infty}\left(\frac{\log{\left(x! \right)}}{x}\right)$$
=
$$\lim_{x \to \infty}\left(\frac{\frac{d}{d x} \log{\left(x! \right)}}{\frac{d}{d x} x}\right)$$
=
$$\lim_{x \to \infty}\left(\frac{\Gamma\left(x + 1\right) \operatorname{polygamma}{\left(0,x + 1 \right)}}{x!}\right)$$
=
$$\lim_{x \to \infty}\left(\frac{\Gamma\left(x + 1\right) \operatorname{polygamma}{\left(0,x + 1 \right)}}{x!}\right)$$
=
$$\infty$$
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
$$\lim_{x \to \infty} \log{\left(x! \right)} = \infty$$
and limit for the denominator is
$$\lim_{x \to \infty} x = \infty$$
Let's take derivatives of the numerator and denominator until we eliminate indeterninateness.
$$\lim_{x \to \infty}\left(\frac{\log{\left(x! \right)}}{x}\right)$$
=
$$\lim_{x \to \infty}\left(\frac{\frac{d}{d x} \log{\left(x! \right)}}{\frac{d}{d x} x}\right)$$
=
$$\lim_{x \to \infty}\left(\frac{\Gamma\left(x + 1\right) \operatorname{polygamma}{\left(0,x + 1 \right)}}{x!}\right)$$
=
$$\lim_{x \to \infty}\left(\frac{\Gamma\left(x + 1\right) \operatorname{polygamma}{\left(0,x + 1 \right)}}{x!}\right)$$
=
$$\infty$$
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)
Other limits x→0, -oo, +oo, 1
$$\lim_{x \to \infty}\left(\frac{\log{\left(x! \right)}}{x}\right) = \infty$$
$$\lim_{x \to 0^-}\left(\frac{\log{\left(x! \right)}}{x}\right) = - \gamma$$
More at x→0 from the left
$$\lim_{x \to 0^+}\left(\frac{\log{\left(x! \right)}}{x}\right) = - \gamma$$
More at x→0 from the right
$$\lim_{x \to 1^-}\left(\frac{\log{\left(x! \right)}}{x}\right) = 0$$
More at x→1 from the left
$$\lim_{x \to 1^+}\left(\frac{\log{\left(x! \right)}}{x}\right) = 0$$
More at x→1 from the right
$$\lim_{x \to -\infty}\left(\frac{\log{\left(x! \right)}}{x}\right) = 0$$
More at x→-oo
$$\lim_{x \to 0^-}\left(\frac{\log{\left(x! \right)}}{x}\right) = - \gamma$$
More at x→0 from the left
$$\lim_{x \to 0^+}\left(\frac{\log{\left(x! \right)}}{x}\right) = - \gamma$$
More at x→0 from the right
$$\lim_{x \to 1^-}\left(\frac{\log{\left(x! \right)}}{x}\right) = 0$$
More at x→1 from the left
$$\lim_{x \to 1^+}\left(\frac{\log{\left(x! \right)}}{x}\right) = 0$$
More at x→1 from the right
$$\lim_{x \to -\infty}\left(\frac{\log{\left(x! \right)}}{x}\right) = 0$$
More at x→-oo