Mister Exam
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- Limit of the function:
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Limit of ((-4+3*x)/(2+3*x))^(1+x)/3
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Limit of (-16+2^x)/(-1+5*sqrt(x)*(5-x))
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Limit of (-14+x^2-5*x)/(-6+x+2*x^2)
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Identical expressions
- log(factorial(x))/x^ two
- logarithm of (factorial(x)) divide by x squared
- logarithm of (factorial(x)) divide by x to the power of two
- log(factorial(x))/x2
- logfactorialx/x2
- log(factorial(x))/x²
- log(factorial(x))/x to the power of 2
- logfactorialx/x^2
- log(factorial(x)) divide by x^2
Limit of the function log(factorial(x))/x^2
The solution
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/log(x!)\
lim |-------|
x->oo| 2 |
\ x /
$$\lim_{x \to \infty}\left(\frac{\log{\left(x! \right)}}{x^{2}}\right)$$
Limit(log(factorial(x))/x^2, 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
$$\lim_{x \to \infty} x^{2} = \infty$$
Let's take derivatives of the numerator and denominator until we eliminate indeterninateness.
$$\lim_{x \to \infty}\left(\frac{\log{\left(x! \right)}}{x^{2}}\right)$$
=
$$\lim_{x \to \infty}\left(\frac{\frac{d}{d x} \log{\left(x! \right)}}{\frac{d}{d x} x^{2}}\right)$$
=
$$\lim_{x \to \infty}\left(\frac{\Gamma\left(x + 1\right) \operatorname{polygamma}{\left(0,x + 1 \right)}}{2 x x!}\right)$$
=
$$\lim_{x \to \infty}\left(\frac{\frac{d}{d x} \operatorname{polygamma}{\left(0,x + 1 \right)}}{\frac{d}{d x} \frac{2 x x!}{\Gamma\left(x + 1\right)}}\right)$$
=
$$\lim_{x \to \infty}\left(\frac{\operatorname{polygamma}{\left(1,x + 1 \right)}}{- \frac{2 x x! \operatorname{polygamma}{\left(0,x + 1 \right)}}{\Gamma\left(x + 1\right)} + 2 x \operatorname{polygamma}{\left(0,x + 1 \right)} + \frac{2 x!}{\Gamma\left(x + 1\right)}}\right)$$
=
$$\lim_{x \to \infty}\left(\frac{\frac{d}{d x} \frac{1}{- \frac{2 x x! \operatorname{polygamma}{\left(0,x + 1 \right)}}{\Gamma\left(x + 1\right)} + 2 x \operatorname{polygamma}{\left(0,x + 1 \right)} + \frac{2 x!}{\Gamma\left(x + 1\right)}}}{\frac{d}{d x} \frac{1}{\operatorname{polygamma}{\left(1,x + 1 \right)}}}\right)$$
=
$$\lim_{x \to \infty}\left(- \frac{\left(- \frac{2 x x! \operatorname{polygamma}^{2}{\left(0,x + 1 \right)}}{\Gamma\left(x + 1\right)} + \frac{2 x x! \operatorname{polygamma}{\left(1,x + 1 \right)}}{\Gamma\left(x + 1\right)} + 2 x \operatorname{polygamma}^{2}{\left(0,x + 1 \right)} - 2 x \operatorname{polygamma}{\left(1,x + 1 \right)} + \frac{4 x! \operatorname{polygamma}{\left(0,x + 1 \right)}}{\Gamma\left(x + 1\right)} - 4 \operatorname{polygamma}{\left(0,x + 1 \right)}\right) \operatorname{polygamma}^{2}{\left(1,x + 1 \right)}}{\left(- \frac{2 x x! \operatorname{polygamma}{\left(0,x + 1 \right)}}{\Gamma\left(x + 1\right)} + 2 x \operatorname{polygamma}{\left(0,x + 1 \right)} + \frac{2 x!}{\Gamma\left(x + 1\right)}\right)^{2} \operatorname{polygamma}{\left(2,x + 1 \right)}}\right)$$
=
$$\lim_{x \to \infty}\left(- \frac{\left(- \frac{x x! \operatorname{polygamma}^{2}{\left(0,x + 1 \right)}}{2 \Gamma\left(x + 1\right)} + \frac{x x! \operatorname{polygamma}{\left(1,x + 1 \right)}}{2 \Gamma\left(x + 1\right)} + \frac{x \operatorname{polygamma}^{2}{\left(0,x + 1 \right)}}{2} - \frac{x \operatorname{polygamma}{\left(1,x + 1 \right)}}{2} + \frac{x! \operatorname{polygamma}{\left(0,x + 1 \right)}}{\Gamma\left(x + 1\right)} - \operatorname{polygamma}{\left(0,x + 1 \right)}\right) \operatorname{polygamma}^{2}{\left(1,x + 1 \right)}}{\operatorname{polygamma}{\left(2,x + 1 \right)}}\right)$$
=
$$\lim_{x \to \infty}\left(- \frac{\left(- \frac{x x! \operatorname{polygamma}^{2}{\left(0,x + 1 \right)}}{2 \Gamma\left(x + 1\right)} + \frac{x x! \operatorname{polygamma}{\left(1,x + 1 \right)}}{2 \Gamma\left(x + 1\right)} + \frac{x \operatorname{polygamma}^{2}{\left(0,x + 1 \right)}}{2} - \frac{x \operatorname{polygamma}{\left(1,x + 1 \right)}}{2} + \frac{x! \operatorname{polygamma}{\left(0,x + 1 \right)}}{\Gamma\left(x + 1\right)} - \operatorname{polygamma}{\left(0,x + 1 \right)}\right) \operatorname{polygamma}^{2}{\left(1,x + 1 \right)}}{\operatorname{polygamma}{\left(2,x + 1 \right)}}\right)$$
=
$$0$$
It can be seen that we have applied Lopital's rule (we have taken derivatives with respect to the numerator and denominator) 3 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
$$\lim_{x \to \infty} x^{2} = \infty$$
Let's take derivatives of the numerator and denominator until we eliminate indeterninateness.
$$\lim_{x \to \infty}\left(\frac{\log{\left(x! \right)}}{x^{2}}\right)$$
=
$$\lim_{x \to \infty}\left(\frac{\frac{d}{d x} \log{\left(x! \right)}}{\frac{d}{d x} x^{2}}\right)$$
=
$$\lim_{x \to \infty}\left(\frac{\Gamma\left(x + 1\right) \operatorname{polygamma}{\left(0,x + 1 \right)}}{2 x x!}\right)$$
=
$$\lim_{x \to \infty}\left(\frac{\frac{d}{d x} \operatorname{polygamma}{\left(0,x + 1 \right)}}{\frac{d}{d x} \frac{2 x x!}{\Gamma\left(x + 1\right)}}\right)$$
=
$$\lim_{x \to \infty}\left(\frac{\operatorname{polygamma}{\left(1,x + 1 \right)}}{- \frac{2 x x! \operatorname{polygamma}{\left(0,x + 1 \right)}}{\Gamma\left(x + 1\right)} + 2 x \operatorname{polygamma}{\left(0,x + 1 \right)} + \frac{2 x!}{\Gamma\left(x + 1\right)}}\right)$$
=
$$\lim_{x \to \infty}\left(\frac{\frac{d}{d x} \frac{1}{- \frac{2 x x! \operatorname{polygamma}{\left(0,x + 1 \right)}}{\Gamma\left(x + 1\right)} + 2 x \operatorname{polygamma}{\left(0,x + 1 \right)} + \frac{2 x!}{\Gamma\left(x + 1\right)}}}{\frac{d}{d x} \frac{1}{\operatorname{polygamma}{\left(1,x + 1 \right)}}}\right)$$
=
$$\lim_{x \to \infty}\left(- \frac{\left(- \frac{2 x x! \operatorname{polygamma}^{2}{\left(0,x + 1 \right)}}{\Gamma\left(x + 1\right)} + \frac{2 x x! \operatorname{polygamma}{\left(1,x + 1 \right)}}{\Gamma\left(x + 1\right)} + 2 x \operatorname{polygamma}^{2}{\left(0,x + 1 \right)} - 2 x \operatorname{polygamma}{\left(1,x + 1 \right)} + \frac{4 x! \operatorname{polygamma}{\left(0,x + 1 \right)}}{\Gamma\left(x + 1\right)} - 4 \operatorname{polygamma}{\left(0,x + 1 \right)}\right) \operatorname{polygamma}^{2}{\left(1,x + 1 \right)}}{\left(- \frac{2 x x! \operatorname{polygamma}{\left(0,x + 1 \right)}}{\Gamma\left(x + 1\right)} + 2 x \operatorname{polygamma}{\left(0,x + 1 \right)} + \frac{2 x!}{\Gamma\left(x + 1\right)}\right)^{2} \operatorname{polygamma}{\left(2,x + 1 \right)}}\right)$$
=
$$\lim_{x \to \infty}\left(- \frac{\left(- \frac{x x! \operatorname{polygamma}^{2}{\left(0,x + 1 \right)}}{2 \Gamma\left(x + 1\right)} + \frac{x x! \operatorname{polygamma}{\left(1,x + 1 \right)}}{2 \Gamma\left(x + 1\right)} + \frac{x \operatorname{polygamma}^{2}{\left(0,x + 1 \right)}}{2} - \frac{x \operatorname{polygamma}{\left(1,x + 1 \right)}}{2} + \frac{x! \operatorname{polygamma}{\left(0,x + 1 \right)}}{\Gamma\left(x + 1\right)} - \operatorname{polygamma}{\left(0,x + 1 \right)}\right) \operatorname{polygamma}^{2}{\left(1,x + 1 \right)}}{\operatorname{polygamma}{\left(2,x + 1 \right)}}\right)$$
=
$$\lim_{x \to \infty}\left(- \frac{\left(- \frac{x x! \operatorname{polygamma}^{2}{\left(0,x + 1 \right)}}{2 \Gamma\left(x + 1\right)} + \frac{x x! \operatorname{polygamma}{\left(1,x + 1 \right)}}{2 \Gamma\left(x + 1\right)} + \frac{x \operatorname{polygamma}^{2}{\left(0,x + 1 \right)}}{2} - \frac{x \operatorname{polygamma}{\left(1,x + 1 \right)}}{2} + \frac{x! \operatorname{polygamma}{\left(0,x + 1 \right)}}{\Gamma\left(x + 1\right)} - \operatorname{polygamma}{\left(0,x + 1 \right)}\right) \operatorname{polygamma}^{2}{\left(1,x + 1 \right)}}{\operatorname{polygamma}{\left(2,x + 1 \right)}}\right)$$
=
$$0$$
It can be seen that we have applied Lopital's rule (we have taken derivatives with respect to the numerator and denominator) 3 time(s)
Other limits x→0, -oo, +oo, 1
$$\lim_{x \to \infty}\left(\frac{\log{\left(x! \right)}}{x^{2}}\right) = 0$$
$$\lim_{x \to 0^-}\left(\frac{\log{\left(x! \right)}}{x^{2}}\right) = \infty$$
More at x→0 from the left
$$\lim_{x \to 0^+}\left(\frac{\log{\left(x! \right)}}{x^{2}}\right) = -\infty$$
More at x→0 from the right
$$\lim_{x \to 1^-}\left(\frac{\log{\left(x! \right)}}{x^{2}}\right) = 0$$
More at x→1 from the left
$$\lim_{x \to 1^+}\left(\frac{\log{\left(x! \right)}}{x^{2}}\right) = 0$$
More at x→1 from the right
$$\lim_{x \to -\infty}\left(\frac{\log{\left(x! \right)}}{x^{2}}\right) = 0$$
More at x→-oo
$$\lim_{x \to 0^-}\left(\frac{\log{\left(x! \right)}}{x^{2}}\right) = \infty$$
More at x→0 from the left
$$\lim_{x \to 0^+}\left(\frac{\log{\left(x! \right)}}{x^{2}}\right) = -\infty$$
More at x→0 from the right
$$\lim_{x \to 1^-}\left(\frac{\log{\left(x! \right)}}{x^{2}}\right) = 0$$
More at x→1 from the left
$$\lim_{x \to 1^+}\left(\frac{\log{\left(x! \right)}}{x^{2}}\right) = 0$$
More at x→1 from the right
$$\lim_{x \to -\infty}\left(\frac{\log{\left(x! \right)}}{x^{2}}\right) = 0$$
More at x→-oo