Mister Exam
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How to use it?
- Limit of the function:
- Limit of factorial(n)^2/factorial(2*n)
-
Limit of (9-x+2*x^2)/(5-x)
-
Limit of ((1+x)^4-(-1+x)^4)/((1+x)^3+(-1+x)^3)
-
Limit of (-4+x^2)/(-8+x^3)
-
Identical expressions
- factorial(n)^ two /factorial(two *n)
- factorial(n) squared divide by factorial(2 multiply by n)
- factorial(n) to the power of two divide by factorial(two multiply by n)
- factorial(n)2/factorial(2*n)
- factorialn2/factorial2*n
- factorial(n)²/factorial(2*n)
- factorial(n) to the power of 2/factorial(2*n)
- factorial(n)^2/factorial(2n)
- factorial(n)2/factorial(2n)
- factorialn2/factorial2n
- factorialn^2/factorial2n
- factorial(n)^2 divide by factorial(2*n)
Limit of the function factorial(n)^2/factorial(2*n)
The solution
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/ 2 \
| n! |
lim |------|
n->oo\(2*n)!/
$$\lim_{n \to \infty}\left(\frac{n!^{2}}{\left(2 n\right)!}\right)$$
Limit(factorial(n)^2/factorial(2*n), n, oo, dir='-')
Lopital's rule
We have indeterminateness of type
i.e. limit for the numerator is
$$\lim_{n \to \infty} n!^{2} = \infty$$
and limit for the denominator is
$$\lim_{n \to \infty} \left(2 n\right)! = \infty$$
Let's take derivatives of the numerator and denominator until we eliminate indeterninateness.
$$\lim_{n \to \infty}\left(\frac{n!^{2}}{\left(2 n\right)!}\right)$$
=
$$\lim_{n \to \infty}\left(\frac{\frac{d}{d n} n!^{2}}{\frac{d}{d n} \left(2 n\right)!}\right)$$
=
$$\lim_{n \to \infty}\left(\frac{n! \Gamma\left(n + 1\right) \operatorname{polygamma}{\left(0,n + 1 \right)}}{\Gamma\left(2 n + 1\right) \operatorname{polygamma}{\left(0,2 n + 1 \right)}}\right)$$
=
$$\lim_{n \to \infty}\left(\frac{n! \Gamma\left(n + 1\right) \operatorname{polygamma}{\left(0,n + 1 \right)}}{\Gamma\left(2 n + 1\right) \operatorname{polygamma}{\left(0,2 n + 1 \right)}}\right)$$
=
$$0$$
It can be seen that we have applied Lopital's rule (we have taken derivatives with respect to the numerator and denominator) 1 time(s)
oo/oo,
i.e. limit for the numerator is
$$\lim_{n \to \infty} n!^{2} = \infty$$
and limit for the denominator is
$$\lim_{n \to \infty} \left(2 n\right)! = \infty$$
Let's take derivatives of the numerator and denominator until we eliminate indeterninateness.
$$\lim_{n \to \infty}\left(\frac{n!^{2}}{\left(2 n\right)!}\right)$$
=
$$\lim_{n \to \infty}\left(\frac{\frac{d}{d n} n!^{2}}{\frac{d}{d n} \left(2 n\right)!}\right)$$
=
$$\lim_{n \to \infty}\left(\frac{n! \Gamma\left(n + 1\right) \operatorname{polygamma}{\left(0,n + 1 \right)}}{\Gamma\left(2 n + 1\right) \operatorname{polygamma}{\left(0,2 n + 1 \right)}}\right)$$
=
$$\lim_{n \to \infty}\left(\frac{n! \Gamma\left(n + 1\right) \operatorname{polygamma}{\left(0,n + 1 \right)}}{\Gamma\left(2 n + 1\right) \operatorname{polygamma}{\left(0,2 n + 1 \right)}}\right)$$
=
$$0$$
It can be seen that we have applied Lopital's rule (we have taken derivatives with respect to the numerator and denominator) 1 time(s)
Other limits n→0, -oo, +oo, 1
$$\lim_{n \to \infty}\left(\frac{n!^{2}}{\left(2 n\right)!}\right) = 0$$
$$\lim_{n \to 0^-}\left(\frac{n!^{2}}{\left(2 n\right)!}\right) = 1$$
More at n→0 from the left
$$\lim_{n \to 0^+}\left(\frac{n!^{2}}{\left(2 n\right)!}\right) = 1$$
More at n→0 from the right
$$\lim_{n \to 1^-}\left(\frac{n!^{2}}{\left(2 n\right)!}\right) = \frac{1}{2}$$
More at n→1 from the left
$$\lim_{n \to 1^+}\left(\frac{n!^{2}}{\left(2 n\right)!}\right) = \frac{1}{2}$$
More at n→1 from the right
$$\lim_{n \to -\infty}\left(\frac{n!^{2}}{\left(2 n\right)!}\right) = \left(-\infty\right)!$$
More at n→-oo
$$\lim_{n \to 0^-}\left(\frac{n!^{2}}{\left(2 n\right)!}\right) = 1$$
More at n→0 from the left
$$\lim_{n \to 0^+}\left(\frac{n!^{2}}{\left(2 n\right)!}\right) = 1$$
More at n→0 from the right
$$\lim_{n \to 1^-}\left(\frac{n!^{2}}{\left(2 n\right)!}\right) = \frac{1}{2}$$
More at n→1 from the left
$$\lim_{n \to 1^+}\left(\frac{n!^{2}}{\left(2 n\right)!}\right) = \frac{1}{2}$$
More at n→1 from the right
$$\lim_{n \to -\infty}\left(\frac{n!^{2}}{\left(2 n\right)!}\right) = \left(-\infty\right)!$$
More at n→-oo