(more precisely, the largest $A > 0$ and the smallest $b>0$) and Suppose $0 < \varepsilon < 1$.
Statement: The function $\frac{\sin x}{x}$ tends to the limit $1$ as $x \to 0^+$ because, given $\varepsilon > 0$, there exists $d > 0$ depending on $\varepsilon$ such that
$\left|\frac{\sin x}{x} - 1\right| < \varepsilon$
whenever $0 < x < d$.
statement/end.
This is also a practice question for my exam coming up.
Using the $\varepsilon$-$\delta$ definition of a limit I have simply replaced $d$ by $A\varepsilon^b$ in the inequality:
$0<|x-0|< A\varepsilon^b$
$|x|< A\varepsilon^b$
$\ln |x| = b \ln(A\varepsilon)$
And then solving for $A$ and $b$ from there. However, I am pretty sure this is wrong, but I am really struggling to find another way to solve this?
Any direction would be greatly appreciated!
Thanks.