We have to find

\(\displaystyle\lim{x}\to{4}{\left({\left(\sqrt{{x}}-{3}\right)}-\frac{{1}}{{{x}-{4}}}\right)}\)

Note that we can not find the value of limit by plugging x=4, as in that case the denominator become zero. Let us first rationalize numerator.

\(((\sqrt x-3)-1)/(x-4)=((\sqrt x-3)-1(\sqrt x-3)+1)/((x-4)(\sqrt x-3)+1) =(((\sqrt x-3)^2)-1^2)/((x-4)(\sqrt x-3)+1) =(x-4)/((x-4)(\sqrt x-3)+1) =1/(\sqrt x-3)+1\)

Therefore we have \(\displaystyle\lim{x}\to{4}\frac{{{\left(\sqrt{{x}}-{3}\right)}-{1}}}{{{x}-{4}}}=\lim{x}\to{4}\frac{{{\left(\sqrt{{x}}-{3}\right)}-{1}}}{{{x}-{4}}}=\lim{x}\to{4}{\left(\frac{{1}}{{\sqrt{{x}}-{3}}}-{1}\right)}=\frac{{1}}{{\sqrt{{4}}-{3}}}+{1}=\frac{{1}}{{2}}\)