I need help please and thanks

Please calculate this limit <br>please help me

Tasya [4]

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:

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

7 0

2 years ago

1.Arsenic-74 is used to locate brain tumors. It has a half-life of 17.5 days. 90 mg wereused in a procedure. Write an equation t

Sophie [7]

$1)\text{ }N_t\text{ = 90\lparen}\frac{1}{2})^{\frac{t}{17.5}}$

2) 5.625 mg will be left

Explanation:

1) Half-life = 17.5 days

initial amount of Arsenic-74 = 90 mg

To get the equation, we will use the equation of half-life:

$\begin{gathered} N_t\text{ = N}_0(\frac{1}{2})^{\frac{t}{t_{\frac{1}{2}}}} \\ where\text{ N}_t\text{ = amount remaining} \\ N_0\text{ = initial amount} \\ t_{\frac{1}{2}\text{ }}\text{ = half-life} \end{gathered}$ $N_t\text{ = 90\lparen}\frac{1}{2})^{\frac{t}{17.5}}$

2) we need to find the remaining amount of Arsenic-74 after 70 days

t = 70

$\begin{gathered} N_t=\text{ 90\lparen}\frac{1}{2})^{\frac{70}{17.5}} \\ N_t\text{ = 5.625 mg} \end{gathered}$

So after 70 days, 5.625 mg will be left

4 0

1 year ago

Factor 2m^3-12m^2+18m completely

Nina [5.8K]

Answer:

The answer is A.

Step-by-step explanation:

Firstly, you have to take out the common terms for this expression. In this expression, the common terms ard 2 and m :

$2 {m}^{3} - 12 {m}^{2} + 18m$

$= 2( {m}^{3} - 6 {m}^{2} + 9m)$

$= 2m( {m}^{2} - 6m + 9)$

Next you have to factorise the brackets :

${m}^{2} - 6m + 9$

$= {m}^{2} - 3m - 3m + 9$

$= m(m - 3) - 3(m - 3)$

$= (m - 3)(m - 3)$

$= {(m - 3)}^{2}$

So the final answer is :

$2m {(m - 2)}^{2}$

6 0

3 years ago

Read 2 more answers

Select the factors of x2 - 4x + 2y - 8

cricket20 [7]

None of the factors shown will give you the above number. In order to get the numbers that are on top, you would need two x's being multiplied by each other to get the x^2 value. Since none of the options have that, it is impossible to achieve that.

6 0

3 years ago

Engineers must design a process for producing ammonia in amounts large enough to ensure a reasonable profit. Which criterion wou

ki77a [65]

The process needs to meet the following criterion:The process must make large amounts of product at a low cost. That is option A.

<h3>What is ammonia?</h3>

Ammonia is defined as a gas compound that is made up of nitrogen and oxygen in the ratio of 1:3 respectively.

The industrial and modern method constructed by Engineers for ammonia production is called steam reforming.

The natural resources used for industrial production of ammonia can be gotten from natural gas such as methane.

To make profit, natural products are used which costs less to acquire but will lead to the production of large ammonia quantity.

Learn more about ammonia here:

brainly.com/question/14445062

#SPJ1

5 0

2 years ago