If f(a) = 2, f'(a) = 1, g(a) = –1, g'(a) = 2 then Ltx→agx.fa-gafxx-a =

fa=2,f1a=1,ga=−1, g1a=−2Ltx→agxfa−gafxx−a00 formLtx→ag'xfa−gaf'x1by L'hospital rule=g'afa−gaf'a=22−−11=4+1=5

If f(a) = 2, f'(a) = 1, g(a) = –1, g'(a) = 2 then Ltx→agx.fa-gafxx-a =

fa=2,f1a=1,ga=−1, g1a=−2Ltx→agxfa−gafxx−a00 formLtx→ag'xfa−gaf'x1by L'hospital rule=g'afa−gaf'a=22−−11=4+1=5

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If f(a) = 2, f'(a) = 1, g(a) = –1, g'(a) = 2 then $\underset{x \to a}{L t} \frac{g (x) . f (a) - g (a) f (x)}{x - a}$ =

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$\frac{1}{5}$

- $\frac{1}{5}$

-5

answer is B.

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Detailed Solution

$\begin{array}{l} f (a) = 2, f^{1} (a) = 1, g (a) = - 1, g^{1} (a) = - 2 \\ \underset{x \to a}{L t} \frac{g (x) f (a) - g (a) f (x)}{x - a} (\frac{0}{0} f o r m) \\ \underset{x \to a}{L t} \frac{g' (x) f (a) - g (a) f' (x)}{1} (b y L' h o s p i t a l r u l e) \\ = g' (a) f (a) - g (a) f' (a) \\ = (2) (2) - (- 1) 1 \\ = 4 + 1 = 5 \end{array}$