Find : $\lim\limits_{n\to +\infty}\frac{\left(1+\frac{1}{n^3}\right)^{n^4}}{\left(1+\frac{1}{(n+1)^3}\right)^{(n+1)^4}}$

Find :

$$\lim\limits_{n\to +\infty}\frac{\left(1+\frac{1}{n^3}\right)^{n^4}}{\left(1+\frac{1}{(n+1)^3}\right)^{(n+1)^4}}$$

My attempt : i don't know is correct or no! I use this rule :

$$\lim\limits_{n\to +\infty}(f(x))^{g(x)}=1^{\infty}$$

Then :

$$\lim\limits_{n\to +\infty}(f(x))^{g(x)}=\lim\limits_{n\to +\infty}e^{g(x)(f(x)-1)}$$

So :

$$\lim\limits_{n\to +\infty}\frac{\left(1+\frac{1}{n^3}\right)^{n^4}}{\left(1+\frac{1}{(n+1)^3}\right)^{(n+1)^4}}$$

$$=\lim\limits_{n\to +\infty}\frac{e^{\frac{n^{4}}{n^{3}}}}{e^{\frac{(n+1)^{4}}{(n+1)^{3}}}}$$

$$=\lim\limits_{n\to +\infty}\frac{e^{n}}{e^{n+1}}$$

$$=\frac{1}{e}$$

Is my approach wrong ?

is this called partial limit calculation ?

Solution 1:

HINT:

Use the inequality $$\left(1+\frac{1}{n}\right)^n < e < \left(1+\frac{1}{n}\right)^{n+1}$$ so $$e^{\frac{1}{n+1}}< \left(1+\frac{1}{n}\right)< e^{\frac{1}{n}}$$ We get lower and upper bounds: $$e^{\frac{n^4}{n^3+1}}< \left(1+\frac{1}{n^3}\right)^{n^4}< e^n $$ $$e^{\frac{(n+1)^4}{(n+1)^3 +1}}< \left(1 + \frac{1}{(n+1)^3}\right)^{(n+1)^4} < e^{n+1}$$

Conclude: $$e^{\frac{n^4}{n^3+1}}/e^{n+1}< \ldots < e^n/e^{\frac{(n+1)^4}{(n+1)^3 +1}}$$ so the limit is $\frac{1}{e}$.

Solution 2:

Your approach leads to the right answer, but there may be some questionable moves there. Me, I would apply $\ln$ to the expression and use

$$\tag 1 \ln (1+u) = u +O(u^2)$$

as $u\to 0.$ Thus in our problem we would get

$$n^4\ln(1+1/n^3) - (n+1)^4 \ln (1 +1/(n+1)^3).$$

Apply $(1)$ to get the answer.

Solution 3:

Assuming that you would like to get more than the limit itself, starting with $$a_n=\frac{\left(1+\frac{1}{n^3}\right)^{n^4}}{\left(1+\frac{1}{(n+1)^3}\right)^{(n+1)^4}}$$ take logarithms $$\log(a_n)=n^4\log\left(1+\frac{1}{n^3}\right)-(n+1)^4\log\left(1+\frac{1}{(n+1)^3}\right)$$ Use twice the expansion $$\log(1+x)=x-\frac{x^2}{2}+\frac{x^3}{3}-\frac{x^4}{4}+O\left(x^5\right)$$ and expand to get $$\log(a_n)=-1-\frac{1}{n^3}+\frac{3}{2n^4}+O\left(\frac{1}{n^5}\right)$$ and continue with Taylor series $$a_n=e^{\log(a_n)}=e\left(1-\frac{1}{n^3}+\frac{3}{2n^4}\right)+O\left(\frac{1}{n^5}\right)$$ which shows the limit and how it is approached. Moreover, this gives a shortcut estimation of $a_n$.

Suppose that, using your pocket calculator, you compute $a_3$. You should get $a_3=0.3594$ while the above truncated expansion gives $\frac{53}{54 e}=0.3611$