If Ltx→01+axbx=e4 , where 'a' anb 'b' are different natural numbers then

ltx→01+axbx=e4eltx→01+ax−1bx=e4eltx→0ab=eab=e4ab=4 ⇒a=4,b=1

If Ltx→01+axbx=e4 , where 'a' anb 'b' are different natural numbers then

ltx→01+axbx=e4eltx→01+ax−1bx=e4eltx→0ab=eab=e4ab=4 ⇒a=4,b=1

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$If \underset{x \to 0}{Lt} {(1 + ax)}^{\frac{b}{x}} = e^{4}, where' a' anb' b' are different natural numbers then$

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$a = 4, b = 2$

$a = 8, b = 4$

$a = 16, b = 8$

$a = 4, b = 1$

answer is D.

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$\begin{array}{l} \underset{x \to 0}{l t} {(1 + a x)}^{\frac{b}{x}} = e^{4} \\ e^{\underset{x \to 0}{l t} (1 + a x - 1) \frac{b}{x}} = e^{4} \\ e^{\underset{x \to 0}{l t} a b} = e^{a b} = e^{4} \\ a b = 4 \Rightarrow a = 4, b = 1 \end{array}$

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If Ltx→01+axbx=e4 , where 'a' anb 'b' are different natural numbers then

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$If \underset{x \to 0}{Lt} {(1 + ax)}^{\frac{b}{x}} = e^{4}, where' a' anb' b' are different natural numbers then$