# How can I calculate the limit of this?

What is the limit? $$\lim_{n\rightarrow\infty}\dfrac{3}{(4^n+5^n)^{\frac{1}{n}}}$$

I don't get this limit. Really, I don't know if it has limit.

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Did you mean as $n$ approaches? – Amzoti Feb 22 '13 at 16:16
@Amsoti Yes, thank's – Henfe Feb 22 '13 at 16:18

HINT: \begin{align*} \frac3{(4^n+5^n)^{1/n}}&=\frac3{(4^n+5^n)^{1/n}}\cdot\frac{\left((1/5)^n\right)^{1/n}}{\left((1/5)^n\right)^{1/n}}\\\\ &=\frac3{5\left(1+\left(\frac45\right)^n\right)^{1/n}} \end{align*}

Can you take the limit of that last expression as $n\to\infty$?

Here’s an intuitive way to think about it. When $n$ is very large, $4^n$ is a very small fraction of $5^n$, so $\left(4^n+5^n\right)^{1/n}$ ought to be just a little more than $\left(5^n\right)^{1/n}=5$.

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Note, that the question has been modified, there is no longer a "^1/n" in the enumerator. – k1next Feb 22 '13 at 16:25
@macydanim: Thanks! – Brian M. Scott Feb 22 '13 at 16:28
@BrianM.Scott I'm sorry – Henfe Feb 22 '13 at 16:30
@Henfe: That’s okay; it wasn’t hard to change. – Brian M. Scott Feb 22 '13 at 16:33

Denote the function $$f(n) = \frac{3}{(4^n +5^n)^{\frac{1}{n}}}$$ Recall logarithm is a continuous function, hence denote $$L(f(n)) = \log 3-\frac{\log(4^n +5^n)}{n}\\ \lim_{n \to \infty} L(f(n)) = \log 3 - \lim_{n \to \infty}\frac{\log(4^n +5^n)}{n}=\log 3 - \lim_{n \to \infty} \frac{4^n \log 4 + 5^n \log 5}{4^n + 5^n} \\ =\log 3 - \log 5=\log \bigg(\frac{3}{5} \bigg)$$ I used here L'Hospital's rule and then divided the fraction through $5^n$. Hence $\lim_{n \to \infty}f(n)=\frac{3}{5}$

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