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Limit of the function/ log(factorial(x))/x^2

Limit of the function log(factorial(x))/x^2

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     /log(x!)\
 lim |-------|
x->oo|    2  |
     \   x   /
lim⁡x→∞(log⁡(x!)x2)\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
oo/oo,

i.e. limit for the numerator is
lim⁡x→∞log⁡(x!)=∞\lim_{x \to \infty} \log{\left(x! \right)} = \infty
and limit for the denominator is
lim⁡x→∞x2=∞\lim_{x \to \infty} x^{2} = \infty
Let's take derivatives of the numerator and denominator until we eliminate indeterninateness.
lim⁡x→∞(log⁡(x!)x2)\lim_{x \to \infty}\left(\frac{\log{\left(x! \right)}}{x^{2}}\right)
=
lim⁡x→∞(ddxlog⁡(x!)ddxx2)\lim_{x \to \infty}\left(\frac{\frac{d}{d x} \log{\left(x! \right)}}{\frac{d}{d x} x^{2}}\right)
=
lim⁡x→∞(Γ(x+1)polygamma⁡(0,x+1)2xx!)\lim_{x \to \infty}\left(\frac{\Gamma\left(x + 1\right) \operatorname{polygamma}{\left(0,x + 1 \right)}}{2 x x!}\right)
=
lim⁡x→∞(ddxpolygamma⁡(0,x+1)ddx2xx!Γ(x+1))\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→∞(polygamma⁡(1,x+1)−2xx!polygamma⁡(0,x+1)Γ(x+1)+2xpolygamma⁡(0,x+1)+2x!Γ(x+1))\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→∞(ddx1−2xx!polygamma⁡(0,x+1)Γ(x+1)+2xpolygamma⁡(0,x+1)+2x!Γ(x+1)ddx1polygamma⁡(1,x+1))\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→∞(−(−2xx!polygamma⁡2(0,x+1)Γ(x+1)+2xx!polygamma⁡(1,x+1)Γ(x+1)+2xpolygamma⁡2(0,x+1)−2xpolygamma⁡(1,x+1)+4x!polygamma⁡(0,x+1)Γ(x+1)−4polygamma⁡(0,x+1))polygamma⁡2(1,x+1)(−2xx!polygamma⁡(0,x+1)Γ(x+1)+2xpolygamma⁡(0,x+1)+2x!Γ(x+1))2polygamma⁡(2,x+1))\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→∞(−(−xx!polygamma⁡2(0,x+1)2Γ(x+1)+xx!polygamma⁡(1,x+1)2Γ(x+1)+xpolygamma⁡2(0,x+1)2−xpolygamma⁡(1,x+1)2+x!polygamma⁡(0,x+1)Γ(x+1)−polygamma⁡(0,x+1))polygamma⁡2(1,x+1)polygamma⁡(2,x+1))\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→∞(−(−xx!polygamma⁡2(0,x+1)2Γ(x+1)+xx!polygamma⁡(1,x+1)2Γ(x+1)+xpolygamma⁡2(0,x+1)2−xpolygamma⁡(1,x+1)2+x!polygamma⁡(0,x+1)Γ(x+1)−polygamma⁡(0,x+1))polygamma⁡2(1,x+1)polygamma⁡(2,x+1))\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)
=
00
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)
Rapid solution [src] 
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00
Expand and simplify
Other limits x→0, -oo, +oo, 1
lim⁡x→∞(log⁡(x!)x2)=0\lim_{x \to \infty}\left(\frac{\log{\left(x! \right)}}{x^{2}}\right) = 0
lim⁡x→0−(log⁡(x!)x2)=∞\lim_{x \to 0^-}\left(\frac{\log{\left(x! \right)}}{x^{2}}\right) = \infty
More at x→0 from the left
lim⁡x→0+(log⁡(x!)x2)=−∞\lim_{x \to 0^+}\left(\frac{\log{\left(x! \right)}}{x^{2}}\right) = -\infty
More at x→0 from the right
lim⁡x→1−(log⁡(x!)x2)=0\lim_{x \to 1^-}\left(\frac{\log{\left(x! \right)}}{x^{2}}\right) = 0
More at x→1 from the left
lim⁡x→1+(log⁡(x!)x2)=0\lim_{x \to 1^+}\left(\frac{\log{\left(x! \right)}}{x^{2}}\right) = 0
More at x→1 from the right
lim⁡x→−∞(log⁡(x!)x2)=0\lim_{x \to -\infty}\left(\frac{\log{\left(x! \right)}}{x^{2}}\right) = 0
More at x→-oo
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