LIMITS AT INFINITY WITH VARIABLE N

About "Limits At Infinity With Variable n"

Limits At Infinity With Variable n :

Here we are going to see how to evaluate limits at infinity with variable n.

Limits At Infinity With Variable n - Examples

Question 1 :

Show that lim n-> (1 + 2 + 3 + .........+ n) / (3n2+ 7n +2)  =  1/6

Solution :

=  lim n-> (1 + 2 + 3 + .........+ n) / (3n2+ 7n +2)

By using the formula for n natural numbers.

=  lim n-> (n(n+1)/2) / (3n2+ 7n + 2)

=  lim n-> (n2 + n) / 2(3n2+ 7n + 2)

=  lim n-> (1 + 1/n) / (6 + 14/n + 2/n2

By applying the limit, we get

=  1/6

Hence the value of Show that lim n-> (1 + 2 + 3 + .........+ n) / (3n2+ 7n +2) is 1/6.

Question 2 :

Show that lim n-> (12+22+.........+(3n)2) / (1+2+....+5n)(2n+3)  =  9/25

Solution :

=  lim n-> (12+22+.........+(3n)2) / (1+2+....+5n)(2n+3)

=  lim n->∞ (3n(3n+1)(3n+2)/6) / (5n(5n+1)/2)(2n+3)

=  lim n->∞ (3n(3n+1)(3n+2)/6) / (5n(5n+1)(2n+3)/2)

=  lim n->∞ 3n(2)(3n+1)(3n+2)/(6(5n)(5n+1)(2n+3)

=  lim n->∞ (3n+1)(3n+2) / 5(5n+1)(2n+3)

=  lim n->∞ (1/5)(9n2+9n+2) / (10n2+17n+3)

Let us divide numerator and denominator by n2

=  lim n->(1/5)(9 + 9/n + 2/n2) / (10 + 5/n + 3/n2)

By applying the limit, we get

=  (1/5) (9/10)

=  9/50

Hence the value of lim n-> (12+22+.........+(3n)2) / (1+2+....+5n)(2n+3) is 9/50.

Question 3 :

Show that lim n-> (1/1⋅2 + 1/2⋅3 + 1/3⋅4+....+1/n(n+1))  = 1

Solution :

Let us write

1/1⋅2 as (1/1) - (1/2)  =  (2 - 1)/1⋅2  =  1/1⋅2 ----(1)

1/2⋅3  =  (1/2) - (1/3)  =  (3 - 2)/2⋅3  =  1/2⋅3 ----(2)

1/3⋅4  =  (1/3) - (1/4)  =  (4 - 3)/3⋅4  =  1/3⋅4 ----(3)

(1) + (2) + (3) ==>

=  (1/1)-(1/2)+(1/2)-(1/3)+(1/3)-(1/4)+........(1/n)- (1/(n+1))

=  1 - (1/(n+1))

=  lim n->∞  [1 - (1/(n+1))]

By applying the limit, we get

=  1

Hence the value of  lim n-> (1/1⋅2 + 1/2⋅3 + 1/3⋅4+....+1/n(n+1)) is 1.

Question 4 :

An important problem in fishery science is to estimate the number of fish presently spawning in streams and use this information to predict the number of mature fish or “recruits” that will return to the rivers during the reproductive period. If S is the number of spawners and R the number of recruits, “Beverton-Holt spawner recruit function” is R(S) = S/(αS + where a and b are positive constants. Show that this function predicts approximately constant recruitment when the number of spawners is sufficiently large.

Solution :

When the number of spawners is sufficiently large means S -∞

R(S) = S/(α S +

=  lim S->∞ S/(α S +

Dividing the equation by S, we get

=  lim S->1/(α + /S

By applying the limit, we get

1/α

Question 5 :

A tank contains 5000 litres of pure water. Brine (very salty water) that contains 30 grams of salt per litre of water is pumped into the tank at a rate of 25 litres per minute. The concentration of salt water after t minutes (in grams per litre) is C(t) = 30t/(200 + t).What happens to the concentration as  -∞

Solution :

To find the quantity of concentration as  -∞

C(t) = lim x->∞ 30t/(200 + t)

C(t) = lim x->∞ 30/(200/t + 1)

=  30/1

=  30

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