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Question: Given f¢(2) = 6 and f¢(1) = 4\(\lim _ { h \rightarrow 0 }\) \(\frac{f(2h + 2 + h^{2}) - f(2)}{f(h - ...

Given f¢(2) = 6 and f¢(1) = 4lim⁡h→0\lim _ { h \rightarrow 0 } f(2h+2+h2)−f(2)f(h−h2+1)−f(1)\frac{f(2h + 2 + h^{2}) - f(2)}{f(h - h^{2} + 1) - f(1)} is equal to

A

3/2

B

3

C

5/2

D

–3

Answer

3

Explanation

Solution

lim⁡h→0\lim_{h \rightarrow 0} f(2h+2+h2)−f(2)f(h−h2+1)−f(1)\frac{f(2h + 2 + h^{2}) - f(2)}{f(h - h^{2} + 1) - f(1)}

= lim⁡h→0\lim _ { h \rightarrow 0 } f(2h+2+h2)−f(2)2h+2+h2−2\frac{f(2h + 2 + h^{2}) - f(2)}{2h + 2 + h^{2} - 2}× h(2+h)h(1−h)\frac{h(2 + h)}{h(1 - h)}

× (h−h2+1)−1f(h−h2+1)−f(1)\frac{(h - h^{2} + 1) - 1}{f(h - h^{2} + 1) - f(1)}

= f¢(2) × lim⁡h→0\lim _ { h \rightarrow 0 } 2+h1−h\frac{2 + h}{1 - h}×1f′(1)\frac{1}{f'(1)}= 6 × 2 × 14\frac{1}{4}= 3

Note that L¢ Hospital rule is not applicable in this case.