Problem with solving a specific limit

Question

So, here is the limit $\lim_{n\to\infty}\left(\frac{3n^{2}+4n-5}{3n^{2}-7n+9}\right)^{n}$

I'm not really sure how I should approach this limit. Would really appreciate if someone could say the in general the steps or perhaps even solve it.

Answer 1

Without changing the actual limit we can take the natural log of this limit while exponentiating it so $$\lim_{n\to\infty} \bigg(\frac{3n^2+5n-5}{3n^2-7n+9}\bigg)^n = e^{\lim_{n\to\infty}}{n\ln\bigg(\frac{3n^2+5n-5}{3n^2-7n+9}\bigg)}$$ Now since the exponent is an indeterminate form, namely $$\infty\cdot 0$$ you can rearrange the equation into the form $\frac{0}{0}$ or $\frac{\infty}{\infty}$ and apply L'Hopitals Rule.

Answer 2

Hint: Write the expression as

$$\left(1+\frac{11n-14}{3n^2-7n+9}\right)^n.$$

If we let $u = (11n-14)/(3n^2-7n+9),$ the above becomes

$$[(1+u)^{1/u}]^{nu}.$$

As $n\to \infty, u\to 0.$ Recall $(1+u)^{1/u}\to \text {___}\,$ ...

Answer 3

$$\lim_{n\to\infty}\left(\frac{3n^{2}+4n-5}{3n^{2}-7n+9}\right)^{n}=\lim_{n\to\infty}\left(1+\frac{11n-14}{3n^{2}-7n+9}\right)^{n}=\lim_{n\to\infty}\left(1+\frac{11-14/n}{3n-7+9/n}\right)^{n}$$ then use $$\lim_{n\to\infty}(1+\frac{a}{n})^n=e^a$$ so the limit will become $e^{\frac{11}{3}}$

Answer 4

In the same spirit as previous answers and comments, consider $$A=\left(\frac{3n^{2}+4n-5}{3n^{2}-7n+9}\right)^{n}$$ Take logarithms $$\log(A)=n\log\left(\frac{3n^{2}+4n-5}{3n^{2}-7n+9}\right)=n\log\left(1+\frac{11n-14}{3n^2-7n+9}\right)$$ Now, remembering that, for small $y$, $\log(1+y)\approx y$, then $$\log(A)\approx n\times \frac{11n-14}{3n^2-7n+9}=\frac{11n^2-14n}{3n^2-7n+9}$$ and hence the result for the limit of $\log(A)$ and then for the limit of $A$.