lower bound and infinimum of a function

Question

If $f$ is the function defined by $f(x)=1-\frac{\sin(x)}{x}$ if $x \neq 0$ and $0$ elsewhere, then $\forall \epsilon>0,\inf_{|x|>\epsilon}f(x)>0.$

Attempt: $f$ is a nonnegative function, so the $\inf$ do exist, but what should we to prove it's nonzero, should we consider a lower bound $\delta(\epsilon)$ (depend of $\epsilon$,) such that $\forall x>\epsilon,f(x)>\delta(\epsilon)$? (Note that $f$ is an even function)

Answer 1

Hint: it is clear that the case of interest is when $|x|\leq1$ (why???). Then you can use that $f'(x)>0$ if $\varepsilon<x\leq 1$. Hope that it helps!