What is the limit of the following sequence? $\lim_{n\to\infty} 8^\frac{n+1}{3n+2}$

Question

What is the limit of the following sequence? $$\lim_{n\to\infty} 8^\frac{n+1}{3n+2}$$

I substitute infinity in $n$ and I get infinity + 1 = infinity, 3*infinity+2 = infinity. Infinity over infinity = indeterminate.

Thus, would we have: 8^indeterminate (read as: "8 to the indeterminate power")? Looks funny, but actually I don't know another way to think about it, just applying the properties of infinity. Could you explain me what this all mean, please?

Answer 1

$$\lim_{n\to\infty} 8^\frac{n+1}{3n+2}$$ $$=\lim_{n\to\infty} 8^\frac{1+\frac{1}{n}}{3+\frac{2}{n}}$$ $$= 8^{\lim_{n\to\infty}\frac{1+\frac{1}{n}}{3+\frac{2}{n}}}$$ $$= 8^\frac{1}{3}$$ $$= 2$$

Answer 2

As $f(x)=8^x$ is continuous and the sequence tends to $\frac{1}{3}$, the resulting limit is $2$.

Answer 3

Let $L=\lim_{n\to\infty} 8^\frac{n+1}{3n+2}$ $$ln(L)=ln(\lim_{n\to\infty} 8^\frac{n+1}{3n+2})$$ Note that log of the limit is the limit of the logarithm $$ln(L)=\lim_{n\to\infty} ln(8^\frac{n+1}{3n+2})$$ $$=ln(8)\lim_{n\to\infty} \frac{n+1}{3n+2}$$ Notice the indeteminate form under the limit, so by L'hopitals Rule, $$=ln(8)\lim_{n\to\infty} \frac{1}{3}$$ $$=\frac{1}{3}ln(8)$$ $$=ln(8^\frac{1}{3})=ln(2)$$ All in all, we get $ln(L)=ln(2)\implies L=2$

Other people have provided good suitable answers as well but i thought this was more mathematical i guess. Hope it helps!

Also note that I used several basic properties of logarithms as well as limits to manipulate the function.