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What is the Limit as X approaches infinity 5-X plus cosXoverX?
There are multiple ways to interpret this question-that is, you could mean either (5-x+cosx)/x or 5-x +(cosx/x). The limit of the second option is negative infinity because as x approaches infinity, |cosx/x|≤1, so 5-x+cosx/x is very close to 5-x, and 5-infinity is basically negative infinity. For the first option, we consider that -1≤cosx≤1. This implies that, as x approaches infinity, lim of (5-x-1)/x≤lim of (5-x+cosx)/x≤lim of (5-x+1)/x. Simplifying, we get that, as x approaches infinity, lim of (4-x)/x≤lim of (5-x+cosx)/x≤(6-x)/x. Simplifying our new limits, we get -1≤lim of (5-x+cosx)/x≤1. It is now clear that the limit of (5-x+cosx)/x as x approaches infinity =negative 1.
Why is it said that poisson distribution is a limiting case of binomial distribution?
This browser is totally bloody useless for mathematical display but...The probability function of the binomial distribution is P(X = r) = (nCr)*p^r*(1-p)^(n-r) where nCr =n!/[r!(n-r)!]Let n -> infinity while np = L, a constant, so that p = L/nthenP(X = r) = lim as n -> infinity of n*(n-1)*...*(n-k+1)/r! * (L/n)^r * (1 - L/n)^(n-r)= lim as n -> infinity of {n^r - O[(n)^(k-1)]}/r! * (L^r/n^r) * (1 - L/n)^(n-r)= lim as n -> infinity of 1/r! * (L^r) * (1 - L/n)^(n-r) (cancelling out n^r and removing O(n)^(r-1) as being insignificantly smaller than the denominator, n^r)= lim as n -> infinity of (L^r) / r! * (1 - L/n)^(n-r)Now lim n -> infinity of (1 - L/n)^n = e^(-L)and lim n -> infinity of (1 - L/n)^r = lim (1 - 0)^r = 1lim as n -> infinity of (1 - L/n)^(n-r) = e^(-L)So P(X = r) = L^r * e^(-L)/r! which is the probability function of the Poisson distribution with parameter L.
How do you find the definition of derivative of f of x equals 3x plus 2?
The definition of the derivative, at a point X = x is the limit, as dx tends to 0, of [f(x+dx)-f(x)]/dx. In this case, therefore, it is lim[3*(x+dx)+2 - (3*x+2)]/dx = lim[3x + 3*dx +2 - 3x - 2]/dx = lim[3*dx/dx] = lim[3] = 3.
Lim x2 3x plus 1?
If you mean: Lim(x→2) 3x + 1 Then the answer is 7 If you mean something else, then you'll need to clarify your question.
How do you calculate integrals that go to infinite?
You do what we call an "improper integral". I will denote the integral of f from a to b as intl a-b (f) here. so we define intl a-infinity (f) as lim b->infinity a-b(f) So it is a limit, and just like all other integrals, it may or may not exist (+/- infinity or infinite uncountable oscilations etc.) You have have to prove yourself though about its properties (it's easy since I reduced it to the regular integral) and you will see it's a perfectly fine definition. If you want examples, I have lots, message me.
