Find the limits of the sequence

[Philip Wong]

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>\infty 3n/2n = \infty/\infty =1is this correct?

 

[n.karthick]

Hai, a straightforward look into 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.

 

[HallsofIvy]

But there is NO "n^n" here so that is irrelevant.

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

 

[n.karthick]

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)

 

[Philip Wong]

But there is NO "n^n" here so that is irrelevant.

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

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

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

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

thanks

 

[Mark44]

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

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 \frac{-\frac{3}{(x+3)(x+2)}}{-\frac{1}{x^2}}= \frac{3x^2}{(x+3)(x+2)}

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

thanks