This page helps you learn how to find the limit of a sequence in an analytic way.

[(2n + 1)/(2n - 2)]^{(5n-1)}

when n tends to +∞

lim [1 + 3/(2n-2)]^{(5n-1)} = lim [1 + 1/[(2n - 2)/3]]^{(2n - 2)/(2n - 2)(5n - 1)} =

lim [[1 + 1/[(2n - 2)/3]]^{(2n - 2)/3}]^{3(5n - 1)/(2n - 2)} = e^{15/2}

because

lim (1 + 1/n)^{n} = e

lim [[1 + 1/[(2n - 2)/3]]

because

lim (1 + 1/n)

n[(n + 1)^{1/2} - n^{1/2}]/(n + 1)

when n tends to +∞

lim n[(n+1)^{1/2} - n^{1/2}][(n+1)^{1/2} + n^{1/2}]/[(n+1)([n+1]^{1/2} + n^{1/2})] =

lim n/[(n+1)([n+1]^{1/2} + n^{1/2})] = lim n/[n+1] 1/[n^{1/2}[(1 + 1/n)^{1/2} + 1]] = 0

lim n/[(n+1)([n+1]

(n!)^{2}/n^{2n}

when n tends to +∞

lim [(n!)/n^{n}]^{2} = 0

because of: 1*2*3*…*n < n*n*n*...*n

because of: 1*2*3*…*n < n*n*n*...*n

[(2n^{2} - n)/(2n^{2} + 1)]^{n2}

when n tends to +∞

lim [1 + (-n - 1)/(2n^{2} + 1)]^{n2} =
lim [1 + 1/[(2n^{2} + 1)/(-n - 1)]]^{n2} =

lim [1 + 1/[(2n^{2} + 1)/(-n - 1)]]^{[(2n2 + 1)/(-n - 1)] [(-n - 1)/(2n2 + 1)]n2} =

e^{-∞} = 0

lim [1 + 1/[(2n

e

[(n+1)/(n-2)]^{3n-1}

when n tends to +∞

lim [1 + 3/(n - 2)]^{(3n - 1)} = lim [1 + 1/[(n - 2)/3]]^{(3n - 1)} =

lim [1 + 1/[(n - 2)/3]]^{(n - 2)/3 (3n - 1) 3/(n - 2) } = e^{9}

lim [1 + 1/[(n - 2)/3]]

n^{2/3}/[n+1]

when n tends to +∞

lim [n^{2}/n^{3}]^{1/3}/[1 + 1/n] = 0

ln(n)/n

when n tends to +∞

lim [ln(n) - ln(n - 1)] = 0

n^{1/n}

when n tends to +∞

lim e^{1/n ln(n)} = 1

[ln(n)]^{1/n}

when n tends to +∞

lim e^{1/n ln(ln(n))} = 1

a^{n}/n

when n tends to +∞ and a > 1

lim a^{n} - a^{n-1} = lim a^{n}(1 - 1/a) = +∞

e(1 - 1/(n+1))^{n}

when n tends to +∞

lim e(1 + 1/(-n-1))^{(-n-1)n/(-n-1)} = e e^{-1} = 1

[n! e^{n}/n^{n}]^{1/n}

when n tends to +∞

lim e/n (n!)^{1/n} = 0

because

lim (1*2*...*n)^{1/n} = lim 1^{1/n} 2^{1/n}*...*n^{1/n} = 1

because

lim (1*2*...*n)

when n tends to +∞

lim sin(n)/n = 0

when n tends to +∞

lim arctg(1/n^{1/2})/[1/n^{1/2}] = 1

[n(n - (n^{2} - 1)^{1/2})]^{1/2}

when n tends to +∞

[n(n - (n^{2} - 1)^{1/2})]^{1/2} = [n(n - (n^{2} - 1)^{1/2})]^{1/2} =

n^{1/2}(n - (n^{2} - 1)^{1/2})^{1/2}(n + (n^{2} - 1)^{1/2})^{1/2}/(n + (n^{2} - 1)^{1/2})^{1/2} =

n^{1/2}(n^{2} - (n^{2} - 1)^{1/2})^{1/2}/[n^{1/2}(1 + (1 - 1/n^{2})^{1/2})^{1/2}] =

1/(1 + (1 - 1/n^{2})^{1/2})^{1/2} = 1/2^{1/2}

n

n

1/(1 + (1 - 1/n

when n tends to +∞

(-1)^{(n + 1)} → there is no limit, **however** see cases below

(-1)^{n-1}(2 + 3/n)

when n tends to +∞. Let us consider cases:

-(2 + 3/(2k)) = -2

(2 + 3/(2k+1)) = 2

1 + n/(n + 1) cos(nπ/2)

when n tends to +∞

1 + 2k/(2k + 1) cos(2kπ/2) = 1 + 2k/(2k + 1) cos(kπ) = 1 + 1 = 2

1 + (2k + 1)/(2k + 2) cos((2k+1)π/2) = 1 + (2k + 1)/(2k + 2) cos((k + 1/2)π) = 1 + 0 = 1

(n - 1)/(n + 1) cos(2nπ/3)

when n tends to +∞

(3k - 1)/(3k + 1) cos(2kπ) = 1

3k/(3k + 2) cos((2k + 2/3)π) = -1/2

(-1)^{n}n

when n tends to +∞

(-1)^{n}n = +∞

(-1)^{n}n = -∞

n^{[(-1)nn]}

when n tends to +∞

n^{[(-1)nn]} = + ∞

n^{[(-1)nn]} = 0

1 + n sin(nπ/2)

when n tends to +∞

1 + 2k sin(kπ) = 1

1 + (2k + 1) sin((k + 1/2)π) = + ∞

1 + (2k + 1) sin((k + 1/2)π) = - ∞

n^{2}/[1 + n^{2}] cos(2nπ/3)

when n tends to +∞

9k^{2}/[1 + 9k^{2}] cos(2kπ) = 1

[9k^{2} + 6k + 1]/[9k^{2} + 6k + 2] cos(2kπ + 2π/3) = 1/2

[9k^{2} + 6k + 1]/[9k^{2} + 6k + 2] cos(2kπ + 2π/3) = -1/2

[1 + 2^{[n(-1)n]}]^{1/n}

when n tends to +∞

[1 + 2^{n}]^{1/n} = 2

[1 + 2^{-n}]^{1/n} = 1

cos^{n}(2nπ/3)

when n tends to +∞

cos^{3k}(2kπ) = 1

cos^{3k+1}((k + 1/3)2π) = 0

If you want to find more analytic examples how to calculate limits go to the page function - limits - analytic examples.

To train another analytic examples containing limits calculations go to the page series - convergence - analytic examples.