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Svetach [21]
3 years ago
15

Please calculate this limit please help me​

Mathematics
1 answer:
Tasya [4]3 years ago
7 0

Answer:

We want to find:

\lim_{n \to \infty} \frac{\sqrt[n]{n!} }{n}

Here we can use Stirling's approximation, which says that for large values of n, we get:

n! = \sqrt{2*\pi*n} *(\frac{n}{e} )^n

Because here we are taking the limit when n tends to infinity, we can use this approximation.

Then we get.

\lim_{n \to \infty} \frac{\sqrt[n]{n!} }{n} = \lim_{n \to \infty} \frac{\sqrt[n]{\sqrt{2*\pi*n} *(\frac{n}{e} )^n} }{n} =  \lim_{n \to \infty} \frac{n}{e*n} *\sqrt[2*n]{2*\pi*n}

Now we can just simplify this, so we get:

\lim_{n \to \infty} \frac{1}{e} *\sqrt[2*n]{2*\pi*n} \\

And we can rewrite it as:

\lim_{n \to \infty} \frac{1}{e} *(2*\pi*n)^{1/2n}

The important part here is the exponent, as n tends to infinite, the exponent tends to zero.

Thus:

\lim_{n \to \infty} \frac{1}{e} *(2*\pi*n)^{1/2n} = \frac{1}{e}*1 = \frac{1}{e}

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Use the discriminant to describe the roots of each equation. Then select the best description. 2m2 + 3 = m.
Tpy6a [65]

The roots of the equation , 2m^{2}+3=m, are non real roots.

<h3>Discriminants are what?</h3>

The quadratic formula's discriminant is represented by the square root symbol b^{2}-4ac. If there are two solutions, one solution, or none at all, the discriminant informs us.

As of now, 2m²+3=m we need to find the kind of roots they have.

2m²+3=m

2m²-m+3=0

Here, we can derive the following values from the equation:

a = 2, b = -1, c = 3

An equation's discriminant is stated as:

D = b²-4ac

D = (-1)(-1) - 4(2)(3)

D = 1 - 24

D = -23

Discriminant can determine the kind of roots that the equation contains.

In this instance, the discriminant is below 0.

The equation's roots are complex conjugates when D < 0.

Hence the roots for the equation are non real roots.

To get more information about discriminant follow the link:

https://brainly.in/question/27214502

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1 year ago
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Hope this helped!

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Answer:

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Step-by-step explanation:

I got it right on my test! :)

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