a$)$ Use interval notation to fill in the blanks that follow.

Since $\underset{x4}{lim}\sqrt[\phantom{2}]{x}=q,,$ for all $lon>0,$ there exists $>0$ such that

if xin $q,$ then $f(x)lon$

$($b$)$ Use the graph of $y=\sqrt[\phantom{2}]{x}$ and algebra to find the largest $$ that corresponds to the given value of $lon,$ according to the appropriate formal limit definition. The fina answer does not need to be simplified.

$\underset{x4}{lim}\sqrt[\phantom{2}]{x}=2,lon=1$