Convergence of sequence method, Math behind intuition

Question

Now I want to find convergence of a sequence:

$$ \lim_{n \to \infty} \sqrt[n]{4^n + 5^n}$$

Now I am pretty sure I have solved this using logic on inspection: $4^n \ll 5^n$ as $n\rightarrow\infty$, hence $\sqrt[n]{4^n + 5^n}=\sqrt[n]{ 5^n} = 5$

Now I could go from here with an epsilon delta proof(maybe), but I can't see how I can Mathematically find this value without intuition(that may be wrong).

Any advice would be greatly appreciated!

Note: The title can be edited freely(with relevance of course), I wasn't sure what to label it.

Answer 1

In the same spirit as Fermat, rewrite $$4^n+5^n=5^n \Big(1+(\frac{4}{5})^n\Big)$$ So $$ {\sqrt[n]{4^n + 5^n}}= {\sqrt[n]{5^n(1 + (\frac{4}{5})^n})}=5 \sqrt[n]{1 + (\frac{4}{5})^n}$$ Now, consider the Taylor series of $\sqrt[n]{1+x}$ when $x$ is small; it is given by $$\sqrt[n]{1+x}=1+\frac{x}{n}+O\left(x^2\right)$$ Replace now $x$ by $\Big(\frac{4}{5}\Big)^n$ and you finally end with $${\sqrt[n]{4^n + 5^n}} \simeq 5+\frac{5}{n} \Big(\frac{4}{5}\Big)^n$$

Let us try for a few values of $n$. For $n=10$ the exact value is $5.051256749$ while the approximation is $5.053687091$. For $n=20$ the exact value is $5.002866636$ while the approximation is $5.002882304$.

Answer 2

$$\lim \limits_{n \to \infty} {\sqrt[n]{4^n + 5^n}}=\lim \limits_{n \to \infty} {\sqrt[n]{5^n(1 + (\frac{4}{5})^n})}=5$$ Since $(\frac{4}{5})^{n}$ goes to zero as $n$ tends to infinity. But David Mitra's answer in his comment is more nice!

Answer 3

Another way: log the expression to get $$ \frac{\log (5^n + 4^n)}{n} = \frac{\log 5^n + \log (1+(\frac{4}{5})^n)}{n} \sim \log 5 + \frac{(\frac{4}{5})^n}{n} \to_n \log 5 $$ Now exponentiate and get $5$.