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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?

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    $\begingroup$ Do the limit of the exponent first as a separate problem. Start that by dividing top and bottom by $n$. $\endgroup$ – coffeemath Nov 5 '15 at 14:19
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$$\lim_{n\to\infty} 8^\frac{n+1}{3n+2}$$ $$=\lim_{n\to\infty} 8^\frac{1+\frac{1}{n}}{3+\frac{2}{n}}$$ $$= 8^{\lim_{n\to\infty}\frac{1+\frac{1}{n}}{3+\frac{2}{n}}}$$ $$= 8^\frac{1}{3}$$ $$= 2$$

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  • $\begingroup$ thank you! I really like your answer! So, in the exponent we apply the following algebraic property, right? a/b = a * 1/b. You jumped a passage, namely, n(1+1/n) at numerator and n(3+2/n) at the denominator. Then we simplify n outside the parenthesis (we might do it in the same passage). I think I even do not need your confirmation...now that I am writing is also clearer...well, thank you! $\endgroup$ – Always learning Nov 5 '15 at 15:32
  • $\begingroup$ Yup I jumped that step....Anyways you're welcome...BTW you can accept my answer if you like it...:-) $\endgroup$ – SchrodingersCat Nov 5 '15 at 15:36
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As $f(x)=8^x$ is continuous and the sequence tends to $\frac{1}{3}$, the resulting limit is $2$.

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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.

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