We have [tex]f(7)=8918[/tex]. By the definition of continuity, we require that
[tex]\displaystyle\lim_{x\to7}f(x)=f(7)[/tex]
So for any [tex]\varepsilon>0[/tex], we want to find sufficient [tex]\delta[/tex] for which
[tex]0<|x-7|<\delta\implies|f(x)-8918|<\varepsilon[/tex]
We have
[tex]|5x^4-9x^3+x-7-8918|=|5 x^4 - 9 x^3 + x - 8925|=|x - 7||5 x^3 + 26 x^2 + 182 x + 1275|[/tex]
Suppose we let [tex]0<\delta\le1[/tex]. Then
[tex]|x-7|\le1\implies x\le8[/tex]
so that
[tex]|5 x^3 + 26 x^2 + 182 x + 1275|\le5|x|^3+26|x|^2+182|x|+1275\le6955[/tex]
[tex]\implies|5x^4-9x^3+x-7-8918|\le6955|x-7|<\varepsilon[/tex]
[tex]\implies|x-7|<\dfrac\varepsilon{6955}[/tex]
which suggests that it suffices to choose
[tex]\delta=\min\left\{1,\dfrac\varepsilon{6955}\right\}[/tex]
in order to meet the required condition.
