If we have sequence $(\frac{4^{n}+1}{5^{n}})_{n\in \mathbb{N_{0}}}$, it is easy to calculate limit of it:
$\lim_{n \to \infty} = \frac{4^{n}+1}{5^{n}}=\frac{(\frac{4}{5})^{n}+\frac{1}{5^{n}}}{1}=\frac{0+0}{1}=0$
But how can i prove it with $\epsilon$, if we know:
$(\forall \epsilon>0)(\exists N\in\mathbb{N}) (\forall n \geq N): \left| a_{n}-a\right|< \epsilon$ ?
We should start with $ \left| \frac{4^{n}+1}{5^{n}}-0 \right|=\left| \frac{4^{n}+1}{5^{n}} \right|=$, how should i continue this proof?
Note that $n \in \mathbb{N}$ implies $$ \frac{4^{n}+1}{5^{n}} = (\frac{4}{5})^{n} + (\frac{1}{5})^{n}. $$ Let $\varepsilon > 0$. Note that $(4/5)^{n} < \varepsilon/2$ if $n\log(4/5) < \log (\varepsilon/2)$ with $\log(4/5) < 0$, so $n > \log(\varepsilon/2)/|\log(4/5)|$ implies $(4/5)^{n} < \varepsilon/2$. Note that likewise $n > \log(\varepsilon/2)/|\log(1/5)|$ implies $(1/5)^{n} < \varepsilon/2$. Note that if $n >$ the greater one of two obtained numbers, then we make it.