Show that $\underset{x0}{lim}\frac{5sin(6x)}{4x}=7\mathrm{.}5.$ Do this by first creating a vector $x$ that has the elements $1\mathrm{.}0,0\mathrm{.}1,0\mathrm{.}01,0\mathrm{.}001,$ and $0\mathrm{.}0001.$ Then, create a new vector $y$ in which each element is determined from the elements of $x$ by $\frac{5sin(6x)}{4x}.$ Compare the elements of $y$ with the value $7\mathrm{.}5.$ Use format long to display the numbers.