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

Hence the answer is 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

After having gone through the stuff given above, we hope that the students would have understood, "Limits At Infinity With Variable n"

Apart from the stuff given in "Limits At Infinity With Variable n" if you need any other stuff in math, please use our google custom search here.

Kindly mail your feedback to v4formath@gmail.com

We always appreciate your feedback.

©All rights reserved. onlinemath4all.com

Recent Articles

  1. SAT Math Videos

    May 22, 24 06:32 AM

    sattriangle1.png
    SAT Math Videos (Part 1 - No Calculator)

    Read More

  2. Simplifying Algebraic Expressions with Fractional Coefficients

    May 17, 24 08:12 AM

    Simplifying Algebraic Expressions with Fractional Coefficients

    Read More

  3. The Mean Value Theorem Worksheet

    May 14, 24 08:53 AM

    tutoring.png
    The Mean Value Theorem Worksheet

    Read More