Find the limits of the sequence

  • #1

Homework Statement


(n+3/n+2)n


Homework Equations


By highest power.


The Attempt at a Solution


highest power = 1

((n/n)+(3/n)/(n/n)+(2/n))n/n

both n/n inside the brackets cancel out. The power n/n becomes 1.

(3/n)/(2/n)

3/n * n/2

3n/2n

lim n>[tex]\infty[/tex] 3n/2n = [tex]\infty[/tex]/[tex]\infty[/tex] =1


is this correct?
 

Answers and Replies

  • #2
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Hai, a straightforward look in to the given sequence seems like there is no limit for this sequence, because it can be reduced to n^n as n gets larger proving that there is no limit for the given sequence.
 
  • #3
HallsofIvy
Science Advisor
Homework Helper
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But there is NO "[itex]n^n[/itex]" here so that is irrelevant.

If [itex]y= ((n+2)/(n+3))^n[/itex] then [itex]ln(y)= nln((n+2)/(n+3))[/itex] and that will have the same limit, as n goes to infinity, as the limit of [itex]xln((x+2)/(x+3))[/itex] as x goes to infinity. By L'Hopital's rule, that will have the same limit as
[tex]\frac{-\frac{3}{(x+3)(x+2)}}{-\frac{1}{x^2}}= \frac{3x^2}{(x+3)(x+2)}[/tex]
which is 3. Since the limit of ln(y) is 3, the limit of y itself is [itex]e^3[/itex].
 
  • #4
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Hi sorry, I thought given sequence is (n+3/n+2)^n and not ((n+3)/(n+2))^n, that's why i replied it can be reduced to n^n.
So considering ((n+3)/(n+2))^n
we can write
ln y= n ln ( (n+3) / (n+2)) = ln ( (n+3) / (n+2)) /(1/n)
This gives 0/0 in RHS when n is infinite
Applying L' Hospitals rule differentiating both numerator and denaminator
we get

ln y= (-1/((n+2)(n+3)))/(-1/n^2)

So as n tends to infinity we get

ln y =1

so y= e^1= 2.7183

which is the limit of the given sequence. I have written matlab code also to verify its limit is 2.7183.

clear all
i=linspace(1,1000,100);
y=((i+3)./(i+2)).^i;
plot(i,y)
 
  • #5
But there is NO "[itex]n^n[/itex]" here so that is irrelevant.

If [itex]y= ((n+2)/(n+3))^n[/itex] then [itex]ln(y)= nln((n+2)/(n+3))[/itex] and that will have the same limit, as n goes to infinity, as the limit of [itex]xln((x+2)/(x+3))[/itex] as x goes to infinity. By L'Hopital's rule, that will have the same limit as
[tex]\frac{-\frac{3}{(x+3)(x+2)}}{-\frac{1}{x^2}}= \frac{3x^2}{(x+3)(x+2)}[/tex]
which is 3. Since the limit of ln(y) is 3, the limit of y itself is [itex]e^3[/itex].

can you please explain how did you get from [itex]y= ((n+2)/(n+3))^n[/itex] to [itex]ln(y)= nln((n+2)/(n+3))[/itex]?

I know the L'Hoptial Rule is: f'(x)/g'(x)
but I don't see how can you derivatives it into [tex]\frac{-\frac{3}{(x+3)(x+2)}}{-\frac{1}{x^2}}= \frac{3x^2}{(x+3)(x+2)}[/tex]

also what do you meant by there is no x? I might have type my question from which may lead to such conclusion.
[itex]y= ((n+3)/(n+2))^n[/itex]n=1 to infinity. does this change anything?

thanks
 
  • #6
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can you please explain how did you get from [itex]y= ((n+2)/(n+3))^n[/itex] to [itex]ln(y)= nln((n+2)/(n+3))[/itex]?
Take the natural log of both sides.
I know the L'Hoptial Rule is: f'(x)/g'(x)
but I don't see how can you derivatives it into [tex]\frac{-\frac{3}{(x+3)(x+2)}}{-\frac{1}{x^2}}= \frac{3x^2}{(x+3)(x+2)}[/tex]

also what do you meant by there is no x? I might have type my question from which may lead to such conclusion.
[itex]y= ((n+3)/(n+2))^n[/itex]n=1 to infinity. does this change anything?

thanks
 

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