Limes problem no. 3

[mathwannabe]

Hello everybody

I just need to know if i did the following problem right:

\(\displaystyle \lim_{n \to \infty} \left(\dfrac{n^2}{n^2+2}\right)^{1-3n^2}=?\)

So, this is how I proceeded:
After taking natural log, and using L'Hospitals rule, I ended up with:

\(\displaystyle \lim_{n \to \infty}a_n=e^{(-18)}\)

Can anyone confirm this?

 

[MarkFL]

I get \(\displaystyle e^6\) as follows:

\(\displaystyle \lim_{n\to\infty}\left(\dfrac{n^2}{n^2+2} \right)^{1-3n^2}=L\)

\(\displaystyle \lim_{n\to\infty}\left(\dfrac{n^2}{n^2+2} \right)\left(\dfrac{n^2}{n^2+2} \right)^{-3n^2}=L\)

\(\displaystyle \lim_{n\to\infty}\left(\dfrac{n^2}{n^2+2} \right)\cdot\lim_{n\to\infty}\left(\dfrac{n^2}{n^2+2} \right)^{-3n^2}=L\)

\(\displaystyle 1\cdot\lim_{n\to\infty}\left(\dfrac{n^2}{n^2+2} \right)^{-3n^2}=L\)

Take the natural log of both sides:

\(\displaystyle \ln\left(\lim_{n\to\infty}\left(\dfrac{n^2}{n^2+2} \right)^{-3n^2} \right)=\ln(L)\)

\(\displaystyle \lim_{n\to\infty}\ln\left(\left(\dfrac{n^2}{n^2+2} \right)^{-3n^2} \right)=\ln(L)\)

\(\displaystyle 3\lim_{n\to\infty}n^2\ln\left(\dfrac{n^2+2}{n^2} \right)=\ln(L)\)

\(\displaystyle 3\lim_{n\to\infty}\dfrac{\ln\left(\dfrac{n^2+2}{n^2} \right)}{\dfrac{1}{n^2}}=\ln(L)\)

We have the indeterminate form 0/0, thus application of L'Hôpital's rule gives:

\(\displaystyle 3\lim_{n\to\infty}\dfrac{\dfrac{n^2}{n^2+2}\cdot \dfrac{n^2(2n)-(n^2+2)(2n)}{n^4}}{-\dfrac{2}{n^3}}=\ln(L)\)

\(\displaystyle 3\lim_{n\to\infty}\dfrac{\dfrac{n^2}{n^2+2}\cdot \dfrac{-4}{n^3}}{-\dfrac{2}{n^3}}=\ln(L)\)

\(\displaystyle 6\lim_{n\to\infty}\dfrac{n^2}{n^2+2}=\ln(L)\)

\(\displaystyle 6\cdot1=\ln(L)\)

\(\displaystyle L=e^6\)

 

[soroban]

Hello, mathwannabe!

I agree with MarkFL . . .

You should be familiar with these two theorems:

. . \(\displaystyle \displaystyle\lim_{x\to\infty}\left(1 + \tfrac{1}{n}\right)^n \;=\;e\)

. . \(\displaystyle \displaystyle\lim_{n\to\infty}\left(1 + \tfrac{a}{n}\right)^n \;=\;e^a\)

\(\displaystyle \displaystyle \lim_{n \to \infty} \left(\dfrac{n^2}{n^2+2}\right)^{1-3n^2}\)

We have: .\(\displaystyle \left(\dfrac{n^2}{n^2+2}\right)^{1-3n^2} \;=\;\left(\dfrac{n^2}{n^2+2}\right)\left(\dfrac{ n^2}{ n^2+2}\right)^{-3n^2}\)

. . . . . \(\displaystyle =\;\left(\dfrac{n^2}{n^2+2}\right)\left(\dfrac{n^2+2}{n^2}\right)^{3n^2} \;=\;\dfrac{1}{1+\frac{2}{n^2}}\left(1 + \dfrac{2}{n^2}\right)^{3n^2} \)

. . . . . \(\displaystyle =\;\dfrac{1}{1+\frac{2}{n^2}}\left[\left(1 + \dfrac{2}{n^2}\right)^{n^2}\right]^3 \)


Then: .\(\displaystyle \displaystyle\lim_{n\to\infty}\dfrac{1}{1+\frac{2}{n^2}}\left[\left(1 + \dfrac{2}{n^2}\right)^{n^2}\right]^3 \;=\;\lim_{n\to\infty}\frac{1}{1 + \frac{1}{n^2}} \cdot \left[\lim_{n\to\infty}\left(1 + \frac{2}{n^2}\right)^{n^2}\right]^3 \)

. . . . . \(\displaystyle =\;\;\dfrac{1}{1+0}(e^2)^3 \;\;=\;\;e^6\)