What is the limit of this sequence $a_n = (\frac{2n^2-2}{2n^2+3})^n$?

Question

The problem itself: $a_n = (\frac{2n^2-2}{2n^2+3})^n$

I know that $(1\pm\frac{\alpha}{n})^n=e^\alpha$, I was trying to use this to solve the problem, but I only got this far:

$a_n=(\frac{1-\frac{1}{n^2}}{1+\frac{3}{2n^2}})^n$ from here i could not figure out where to go.

I tried to exponentiate the denominator (the top part) to $\frac{1}{n}$ but that did not get me further. It looked like this: $((1-\frac{1}{n^2})^{n^2})^{1/n}=(e^1)^{1/n}$

No idea what to do, where should I start? I don't even know how to call this type of sequence. Don't know what to search.

Thanks for your time!

EDIT: In the [example where it is solved] the teacher uses order principle to do it. But i do not understand this.

EDIT2: Found a possible solution, it it right?

$a_n=(\frac{1-\frac{1}{n^2}}{1+\frac{3/2}{n^2}})^n=(\frac{e^{-1}}{e^{3/2}})^{1/n}=(e^{5/2})^0=1$

Answer 1

All the answers here, as well as teacher's solution mentioned in OP, make use of the definition of the number $e.$ An elementary solution can be based on the Bernoulli inequality (proved easily by induction) $$(1+x)^n\ge 1+nx,\quad x\ge -1$$ We have $$1\ge \left ({2n^2-2\over 2n^2+3}\right )^n=\left (1-{5\over 2n^2+3}\right )^n\\ \ge 1-{5n\over 2n^2+3}\ge 1-{5\over 2n}$$ By the squeeze theorem the limit is equal $1.$

Answer 2

Using the notable limit

$$\lim_{n \to \infty} \left(1 + \frac{a}{n}\right)^n = e^a $$

we have

$$\left(\frac{2n^2-2}{2n^2+3}\right)^n=\left[\left(1+\frac{-5}{(n^2+3)}\right)^{(n^2+3)}\right]^{\frac n{(n^2+3)}}\to (e^{-5})^0=1$$

Answer 3

$\lim \frac{2 n^2-2}{2 n^2+3}$=1

Proof:

Do a limit expansion at $n=\infty$. This is $1-\frac{5}{2 n^2}+\frac{15}{4 n^4}+O{(n^{-5})}$.

Intermediate step: $\frac{2 n^2-2}{2 n^2+3}=1-\frac{5}{3+2n^2}$. For larger $n$ the denominator addition 3 is small. The next correction is easily calculated. This might too be computed with the Mathematica built-in: Series[(2 n^2-2)/(2 n^2+3),{n,Infinity,4}].

Answer 4

Thanks to everyone. Based on your solutions I've found 2 solutions that I understand:

First:

$$\frac{e^{n \cdot \ln(1+\frac{-1}{n^2})}}{e^{n \cdot \ln(\frac{1+3/2}{n^2})}}=\frac{e^{n \cdot \ln(1)}}{e^{n \cdot \ln(1)}}=e^0=1$$

Second:

$$a_n=\left(\frac{1+\frac{-1}{n^2}}{1+\frac{3/2}{n^2}}\right)^{n^2 \cdot 1/n} =\left(\frac{e^{-1}}{e^{3/2}}\right)^{1/n}=(e^{5/2})^0=1$$

I'm not sure if these are mathematically correct, feel free to edit. Although it seems like they do the job. Thanks for everyone.