lim _{x rightarrow 0}(1+3x)^{frac{2}{x}} = | vedclass

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Question

(B) We know that the standard limit formula is $\lim _{x ightarrow 0}(1+ax)^{\frac{1}{ax}} = e$ for $a 
eq 0$. Given expression is $\lim _{x ight

Vedclass · Accepted Answer

$e^6$

Vedclass · Answer

(B) We know that the standard limit formula is $\lim _{x ightarrow 0}(1+ax)^{\frac{1}{ax}} = e$ for $a 
eq 0$. Given expression is $\lim _{x ightarrow 0}(1+3x)^{\frac{2}{x}}$. We can rewrite the exponent as $\frac{2}{x} = 6 	imes \frac{1}{3x}$. Substituting this into the limit,we get $\lim _{x ightarrow 0}(1+3x)^{\frac{2}{x}} = \lim _{x ightarrow 0} \left((1+3x)^{\frac{1}{3x}}ight)^6$. Using the standard limit property,this expression evaluates to $e^6$.

$\lim _{x \rightarrow 0}(1+3x)^{\frac{2}{x}} = $

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