The limit should be 0.
How about three things that are infinite. Counted number are infinite. The complete statement of PI is infinite. The result of 1 divided by 3 is infinite.
Compare a series to a known series. So take the harmonic series {1/1 + 1/2 + 1/3 + ... + 1/n}, which diverges.For each number n [n>1], LN(n) < n, so 1/(LN(n)) > 1/n. So since each term in 1/LN(n) is greater than each term in the divergent series {1/n}, then the series 1/LN(n) diverges.
Use the product rule.y = x lnxy' = x (ln x)' + x' (ln x) = x (1/x) + 1 ln x = 1 + ln xUse the product rule.y = x lnxy' = x (ln x)' + x' (ln x) = x (1/x) + 1 ln x = 1 + ln xUse the product rule.y = x lnxy' = x (ln x)' + x' (ln x) = x (1/x) + 1 ln x = 1 + ln xUse the product rule.y = x lnxy' = x (ln x)' + x' (ln x) = x (1/x) + 1 ln x = 1 + ln x
ln 1 = 0 e0=1
An infinite number of integers can be divided by 126.
converges to zero (I think)
x^(ln(2)/ln(x)-1)
0.5
1/ln(x)*e^(1/x) if you differentiate e^(1/x), you will get ln(x)*e^(1/x). times this by 1/ln(x) and you get you original equation. Peace
How about three things that are infinite. Counted number are infinite. The complete statement of PI is infinite. The result of 1 divided by 3 is infinite.
3
The limit is 4.
1 time infinity equals infinity. Infinite divided by infinite equals 1. There's your answer. * * * * * Except that it is not true. 1 times infinity is, indeed, infinity. But infinity divided by infinity need not be 1. See for example, the paradox of Hibert's Hotel at the attached link.
The first of an infinite series of solutions is: log10(-2.4969)=ln(-2.4969)/ln(10)=ln(2.4969)/ln(10) +PI*i/ln(10) = .397 + 1.364*i There are an infinite number of additional solutions of the form: .397 + 1.364*i +2*PI*k/ln(10) where k is any integer greater than 0. I got this number by using the change of base identity and a common, complex log identity, neither of which I'm deriving. If you haven't been taught it yet, i = sqrt(-1).
1
There are an infinite number of possible answers. Amongst the simpler is 1428 divided by 1.
Compare a series to a known series. So take the harmonic series {1/1 + 1/2 + 1/3 + ... + 1/n}, which diverges.For each number n [n>1], LN(n) < n, so 1/(LN(n)) > 1/n. So since each term in 1/LN(n) is greater than each term in the divergent series {1/n}, then the series 1/LN(n) diverges.