Given that $Ɛ>0$ find a $>0$ so that
$|x|<⇒|sin^4(\sqrt{|x|})|<Ɛ$
We are also given that
$|sin(x)|<|x|$
Here is my attempt at finding a relationship between $Ɛ$ and $$:
Firstly we can find that $|sin^4(\sqrt{|x|})|≤|sin(\sqrt{|x|})|$ because...
The function $|sin(x)|$ will always yield a decimal from $0$ to $1$ e.g. $0≤|sin(x)|≤1$. This decimal to any power greater than or equal to one will be smaller or equal than the decimal itself. e.g. $0.5^4≤0.5$ and $1^4≤1$
Therefore $|sin^4(\sqrt{|x|})|≤|sin(\sqrt{|x|})|<\sqrt{|x|}$ and $|sin^4(\sqrt{|x|})|<Ɛ$
I would like to assume that $\sqrt{|x|}<Ɛ$ but it's not always true, all we know is $|sin^4(\sqrt{|x|})|<Ɛ$
I think it has something to do with $|x|<$ given at the beginning but I am not sure.
The answer to the relationship is $|x|<=Ɛ^2$
Given $\varepsilon > 0$, set $\delta := \sqrt{\varepsilon}$. Take an $x$ such that $|x| < \delta$. Then $$|\sin^4(\sqrt{|x|})| = |\sin(\sqrt{|x|})| ^4 \leq (\sqrt{|x|})^4 = |x|^2 < \delta^2 = |\varepsilon| = \varepsilon$$ where for the $|\sin(\sqrt{|x|})| ^4 \leq (\sqrt{|x|})^4$ inequality one uses that $|\sin (y)| \leq |y|$ for any real number $y$.