Solve Limit at Infinity Problem: Tips & Help

[Alexstrasuz1]

Hi, I've been doing limit problems, and just got to this problem and I can't solve it. I would love some tips; you don't have to solve my problem.
Screenshot by Lightshot

 

[Ackbach]

What is the $2n+3$ doing? Your formatting is a bit weird. Is the limit
$$\lim_{n\to \infty}\left(1+\frac{3n-1}{n^2+1}\right)^{\!\!2n+3}?$$

 

[Alexstrasuz1]

Im supposed to get it to this lim(1+1/n)^n=e

 

[Siron]

I suggest the following:

Rewrite the exponent $2n+3$ as follows
$$\frac{(2n+3)(3n-1)}{3n-1} = \frac{6n^2+7n-3}{3n-1} = \frac{6(n^2+1)}{3n-1}+\frac{7n-9}{3n-1}$$

Then we can write the limit as
$$\displaystyle \left[\lim_{n \to \infty} \left(1+\frac{1}{\frac{n^2+1}{3n-1}}\right)^{\frac{6(n^2+1)}{3n-1}}\right]\left[\lim_{n \to \infty} \left(1+\frac{1}{\frac{n^2+1}{3n-1}}\right)^{\frac{7n-9}{3n-1}}\right]$$

The left limit is equal to $e^6$, the right limit is equal to $1$. Hence the solution is $e^6$.

 

[Alexstrasuz1]

I solved it by dividing fraction by n and got 3/n and then just did some standard work and got e₆

 

[topsquark]

$$\displaystyle \left[\lim_{n \to \infty} \left(1+\frac{1}{\frac{n^2+1}{3n-1}}\right)^{\frac{6(n^2+1)}{3n-1}}\right]\left[\lim_{n \to \infty} \left(1+\frac{1}{\frac{n^2+1}{3n-1}}\right)^{\frac{7n-9}{3n-1}}\right]$$

The left limit is equal to $e^6$, the right limit is equal to $1$. Hence the solution is $e^6$.

In regard to the second factor, I can intuitively understand what you did, but Mathematically I am lost. How did you know the limit of the second factor is 1?

-Dan

 

[MarkFL]

In regard to the second factor, I can intuitively understand what you did, but Mathematically I am lost. How did you know the limit of the second factor is 1?

-Dan

As $n\to\infty$, the base goes to $1$ and the exponent goes to a finite value (7/3) so we have a determinate form of $1$. :D

 

[topsquark]

Ah! The little details! Thanks. :)

-Dan

 

[Alexstrasuz1]

Hey I solved this on my way and today I showed it to my teacher and she said it isn't right way to solve it even I got the same result
I divided 3n-1/n^2+1 with n so I got 3/n and I wrote it as 1/n/3 and put that n/3=t where n=3t and I got in exponent 6t+3 and got that e^6.

 

[Siron]

Hey I solved this on my way and today I showed it to my teacher and she said it isn't right way to solve it even I got the same result
I divided 3n-1/n^2+1 with n so I got 3/n and I wrote it as 1/n/3 and put that n/3=t where n=3t and I got in exponent 6t+3 and got that e^6.

If I understand you correctly, you mean:
$$\frac{3n-1}{n^2+1} \frac{1}{n} = \frac{3}{n}?$$
The only thing you could do is to divide the numerator and denominator by $n$ but I don't see how that could lead to a solution.