Inequality proof as a part of calculus lesson As part of a calculus lesson I was required to prove that:
(1) if $\ |x-3| < \frac{1}{2},\ $ then $\ \bigg|\displaystyle{\frac{\sin(x^2 -8x+15)}{4x-7}}\bigg| < \frac{1}{2}$
So, by using $|\sin(t)| \le |t|,$ I can prove that:
$$\bigg|\frac{\sin(x^2 -8x+15)}{4x-7}\bigg| \le \frac{5}{6}|x-3|$$
After proving this inequality I assume that the initial requirement is proved, and therefore I'm done, is that it or am I missing something?
Thanks for your assistance :)
 A: In this case, just note $\sin x \leqslant 1$ and $|4x-7| > |4\cdot \frac52-7|=3 $ so $$\left|\frac{\sin(x^2-8x+15)}{4x-7}\right| < \frac13$$
A: Because, $x^{2}-8x+15=(x-3)(x-5)\ $, you can start saying that:
First,
$$ \vert{\sin(x^{2}-8x+15)}\vert\leq {x^{2}-8x+15} $$
and then you start like this 
$$ \bigg\vert\frac{\sin(x^{2}-8x+15)}{4x-7}\bigg\vert=\frac{\vert{\sin(x^{2}-8x+15)}\vert}{\vert{4x-7}\vert}\leq\frac{\vert x^{2}-8x+15\vert}{\vert{4x-7}\vert}=\frac{\vert{(x-3)(x-5)}\vert}{\vert{4x-7}\vert}$$
Suppose that $\ \displaystyle{\vert{x-3}\vert<\frac{1}{2}}\ $, then
$$ -\frac{1}{2}<x-3<\frac{1}{2}$$
$$ 3-\frac{1}{2}<x<3+\frac{1}{2} $$
$$ \frac{5}{2}<x<\frac{7}{2} $$
and with the last condition you can bound $\displaystyle{\frac{1}{\vert{4x-7}\vert}}$ and $\vert{x-5}\vert$. In fact,
$$ \frac{5}{2}<x<\frac{7}{2} \Rightarrow  \frac{5}{2}-5<x-5<\frac{7}{2}-5 $$
this means
$$ -\frac{5}{2}<x-5<-\frac{3}{2} $$
and the absolute value function is decreasing in the negatives real numbers:
$$ \frac{3}{2}<\vert{x-5}\vert<\frac{5}{2}, $$
so you have $$ \vert{x-5}\vert<\frac{5}{2}. $$
Also, 
$$ \frac{5}{2}<x<\frac{7}{2} \Rightarrow 10<4x<14  \Rightarrow 3<4x-7<7$$
so $\vert{4x-7}\vert>3$ and this implies:
$$ \displaystyle{\frac{1}{\vert{4x-7}\vert}}<\frac{1}{3}. $$
Then you can multiply both inequalities below:
$$ \vert{x-5}\vert<\frac{5}{2} \qquad \displaystyle{\frac{1}{\vert{4x-7}\vert}}<\frac{1}{3}$$
to get:
$$ \frac{\vert{x-5}\vert}{\vert{4x-7}\vert}<\frac{5}{6}. $$
So,
$$ \bigg\vert\frac{\sin(x^{2}-8x+15)}{4x-7}\bigg\vert\leq\frac{\vert x^{2}-8x+15\vert}{\vert{4x-7}\vert}=\frac{\vert{(x-3)(x-5)}\vert}{\vert{4x-7}\vert}<\frac{5}{6}\vert{x-3}\vert<\frac{5}{6}\cdot\frac{1}{2}=\frac{5}{12}<\frac{1}{2}.$$
A: HINT.-You want to have $\dfrac52\lt x\lt \dfrac 72$ and your function $|f(x)|=\ \bigg|\displaystyle{\frac{\sin(x^2 -8x+15)}{4x-7}}\bigg| < \frac{1}{2}$ is 
decreasing in this interval from $|f(\frac 52)|=|0.39952128217|$ till
$|f(\frac 72)|=|-0.102734043987|$
