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Q)

$\lim\limits_{h\to 0}\large\frac{f(2h+2+h^2)-f(2)}{f(h-h^2+1)-f(1)}$ given that $f'(2)=6$ and $f'(1)=4$

$\begin{array}{1 1}(a)\;does\;not\;exist&(b)\;is\;equal\;to\;-3/2\\(c)\;is\;equal\;to\;3/2&(d)\;is\;equal\;to\;3\end{array}$

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A)
$\lim\limits_{h\to 0}\large\frac{f(2h+2+h^2)-f(2)}{f(h-h^2+1)-f(1)}$
$\Rightarrow \lim\limits_{h\to 0}\large\frac{f'(2h+2+h^2)(2+2h)}{f'(h-h^2+1)(1-2h)}$
$\Rightarrow \large\frac{6\times 2}{4\times 1}$
$\Rightarrow \large\frac{3}{2}$
Hence (c) is the correct answer.
(6x2)/4x1 = 12/4 = 3 not 3/2
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