First, we simplify the denominator;
1 ( 2 + h ) 2 − 1 4 = 4 − ( 2 + h ) 2 4 ( 2 + h ) 2 = 4 − ( h 2 + 4 h + 4 ) 4 ( 2 + h ) 2 = − ( h 2 + 4 h ) ) 4 ( 2 + h ) 2 = − h ( h + 4 ) 4 ( 2 + h ) 2 {\displaystyle {\frac {1}{(2+h)^{2}}}-{\frac {1}{4}}={\frac {4-(2+h)^{2}}{4(2+h)^{2}}}={\frac {4-(h^{2}+4h+4)}{4(2+h)^{2}}}={\frac {-(h^{2}+4h))}{4(2+h)^{2}}}={\frac {-h(h+4)}{4(2+h)^{2}}}}
Now plug this back to the limit expression and then we obtain the limit as
lim h → 0 1 ( 2 + h ) 2 − 1 4 h = lim h → 0 1 ( 2 + h ) 2 − 1 4 ⋅ 1 h = lim h → 0 − h ( h + 4 ) 4 ( 2 + h ) 2 ⋅ 1 h = lim h → 0 − ( h + 4 ) 4 ( 2 + h ) 2 ⋅ = − 4 4 ⋅ 2 2 = − 1 4 . {\displaystyle {\begin{aligned}\lim _{h\to 0}{\frac {{\frac {1}{(2+h)^{2}}}-{\frac {1}{4}}}{h}}&=\lim _{h\to 0}{{\frac {1}{(2+h)^{2}}}-{\frac {1}{4}}}\cdot {\frac {1}{h}}\\&=\lim _{h\to 0}{\frac {-h(h+4)}{4(2+h)^{2}}}\cdot {\frac {1}{h}}=\lim _{h\to 0}{\frac {-(h+4)}{4(2+h)^{2}}}\cdot =-{\frac {4}{4\cdot 2^{2}}}=\color {blue}-{\frac {1}{4}}.\end{aligned}}}
