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trasher [3.6K]
3 years ago
11

For what value of c is the relation a function (2,8),(12,3),(c,4),(-1,8),(0,3)

Mathematics
1 answer:
Fittoniya [83]3 years ago
3 0
The value of c is { c | c ∈ R, c ∉ {-1,0,2,12} }.
c is a number such that it is not in the set {-1,0,2,12} else we end up with a domain value having multiple range values, contradicting with the definition of a function.


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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.

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

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