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

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?

• Do the limit of the exponent first as a separate problem. Start that by dividing top and bottom by $n$. – coffeemath Nov 5 '15 at 14:19

$$\lim_{n\to\infty} 8^\frac{n+1}{3n+2}$$ $$=\lim_{n\to\infty} 8^\frac{1+\frac{1}{n}}{3+\frac{2}{n}}$$ $$= 8^{\lim_{n\to\infty}\frac{1+\frac{1}{n}}{3+\frac{2}{n}}}$$ $$= 8^\frac{1}{3}$$ $$= 2$$
As $f(x)=8^x$ is continuous and the sequence tends to $\frac{1}{3}$, the resulting limit is $2$.
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 $ln(L)=ln(2)\implies L=2$