Find the limits of the sequence

• Philip Wong
In summary, the sequence (n+3/n+2)^n can be rewritten as ((n+3)/(n+2))^n. Using L'Hopital's rule, we can take the natural log of both sides and find the limit of the sequence as n approaches infinity. Taking the derivative of the numerator and denominator, we get ln(y)= nln((n+2)/(n+3)), which simplifies to -3x^2/(x+3)(x+2). As n approaches infinity, this limit becomes 3, and since the limit of ln(y) is 3, the limit of y itself is e^3. This is verified by a MATLAB code, and it is important to note that
Philip Wong

(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?

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.

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 $e^3$.

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)

HallsofIvy said:
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.
$y= ((n+3)/(n+2))^n$n=1 to infinity. does this change anything?

thanks

Philip Wong said:
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.
Philip Wong said:
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.
$y= ((n+3)/(n+2))^n$n=1 to infinity. does this change anything?

thanks

What is a sequence and how do you find its limits?

A sequence is a list of numbers that follow a specific pattern. To find the limit of a sequence, you must determine the value that the sequence approaches as more terms are added.

What are the different types of sequences and how do you find their limits?

There are three types of sequences: convergent, divergent, and oscillating. To find the limit of a convergent sequence, you can use the limit laws or the squeeze theorem. For divergent sequences, there is no limit. And for oscillating sequences, you must determine the value that the sequence oscillates around.

How do you know if a sequence has a limit?

A sequence has a limit if the terms of the sequence get closer and closer to a single value as more terms are added. This means that the sequence is convergent and has a finite limit.

Can the limit of a sequence be infinity?

Yes, the limit of a sequence can be infinity if the terms of the sequence increase without bound. This means that the sequence is divergent and does not have a finite limit.

What are some common methods for finding the limits of sequences?

Some common methods for finding the limits of sequences include using the limit laws, the squeeze theorem, and determining the behavior of the sequence as the number of terms approaches infinity. Graphing the sequence can also be helpful in visualizing its behavior and finding its limit.

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