SOLUTION
From the limit definition, we have that
[tex]f^{\prime}(x)=\lim _{h\to0}\frac{f(x+h)-f(x)}{h}[/tex]Now applying we have
[tex]\begin{gathered} f\mleft(x\mright)=x^2-5 \\ f^{\prime}(x)=\lim _{h\to0}\frac{f(x+h)-f(x)}{h} \\ =\lim _{h\to0}\frac{((x+h)^2-5)-(x^2-5)}{h} \\ =\lim _{h\to0}\frac{x^2+2xh+h^2^{}-5-(x^2-5)}{h} \\ =\lim _{h\to0}\frac{x^2+2xh+h^2-5-x^2+5}{h} \\ =\lim _{h\to0}\frac{x^2-x^2+2xh+h^2-5+5}{h} \\ =\lim _{h\to0}\frac{2xh+h^2}{h} \end{gathered}[/tex]factorizing for h, we have
[tex]\begin{gathered} =\lim _{h\to0}\frac{h(2x+h)^{}}{h} \\ \text{cancelling h} \\ =\lim _{h\to0}2x+h \\ =2x \end{gathered}[/tex]So, when x = 3, we have
[tex]\begin{gathered} =2x \\ =2\times3 \\ =6 \end{gathered}[/tex]Hence, the answer is 6
