If f1=3,f'1=6 then Ltx→0f1+xf11x=

Ltx→0f1+xf11/x=Ltex→0f1+x−f1x⋅f1 1∞ model =Ltex→0f'1+xf1=ef'1f1 by LH rule =e63=e2

If f1=3,f'1=6 then Ltx→0f1+xf11x=

Ltx→0f1+xf11/x=Ltex→0f1+x−f1x⋅f1 1∞ model =Ltex→0f'1+xf1=ef'1f1 by LH rule =e63=e2

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If $f (1) = 3, f' (1) = 6$ then $\underset{x \to 0}{L t} {[\frac{f (1 + x)}{f (1)}]}^{\frac{1}{x}} =$

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$\sqrt{e}$

$e^{2}$

answer is D.

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Detailed Solution

$\underset{x \to 0}{L t} {[\frac{f (1 + x)}{f (1)}]}^{1 / x} = \underset{e^{x \to 0}}{L t} \frac{f (1 + x) - f (1)}{x \cdot f (1)} (1^{\infty} m o d e l) = \underset{e^{x \to 0}}{L t} \frac{f' (1 + x)}{f (1)} = e^{\frac{f' (1)}{f (1)}} b y L H r u l e = e^{\frac{6}{3}} = e^{2}$