limh→0 f2h+2+h2−f(2)fh−h2+1−f(1) = [given that f'(2)=6 and f′(1)=4]

limh→0 f2h+2+h2−f(2)fh−h2+1−f(1)=limh→0 f2h+2+h2−f(2)(2+2h)×(2+2h)fh−h2+1−f(1)(1−2h)×(1−2h)=limh→0 f′2h+2+h2(2+2h)f′h−h2+1(1−2h)=f′(2)f′(1)⋅21=6×24×1=3

limh→0 f2h+2+h2−f(2)fh−h2+1−f(1) = [given that f'(2)=6 and f′(1)=4]

limh→0 f2h+2+h2−f(2)fh−h2+1−f(1)=limh→0 f2h+2+h2−f(2)(2+2h)×(2+2h)fh−h2+1−f(1)(1−2h)×(1−2h)=limh→0 f′2h+2+h2(2+2h)f′h−h2+1(1−2h)=f′(2)f′(1)⋅21=6×24×1=3

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$lim_{h \to 0} \frac{f (2 h + 2 + h^{2}) - f (2)}{f (h - h^{2} + 1) - f (1)}$ = [given that f'(2)=6 and $f^{'} (1) = 4$ ]

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Detailed Solution

$\begin{array}{l} lim_{h \to 0} \frac{f (2 h + 2 + h^{2}) - f (2)}{f (h - h^{2} + 1) - f (1)} \\ = lim_{h \to 0} \frac{\frac{f (2 h + 2 + h^{2}) - f (2)}{(2 + 2 h)} \times (2 + 2 h)}{\frac{f (h - h^{2} + 1) - f (1)}{(1 - 2 h)} \times (1 - 2 h)} \\ = lim_{h \to 0} \frac{f^{'} (2 h + 2 + h^{2}) (2 + 2 h)}{f^{'} (h - h^{2} + 1) (1 - 2 h)} \\ = \frac{f^{'} (2)}{f^{'} (1)} \cdot \frac{2}{1} = \frac{6 \times 2}{4 \times 1} = 3 \end{array}$