If $\lim _{x \rightarrow 3}\left(\frac{x^n-3^n}{x-3}\right)=108$ and $n \in \mathbf{N}$, then the value of ' $n$ ' is

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Aicharya · Accepted Answer

$\lim _{x ightarrow 3} \frac{x^n-3^n}{x-3}=108$ LHS is in $\div$ from, so apply L-Hospital rule $ \begin{aligned} \lim _{x ightarrow 3} \frac{n \cdot x^{n-1}-0}{1-0} & =108 \ \lim _{x ightarrow 3} n \cdot x^{n-1} & =108 \ n \cdot 3^{n-1} & =108 \ n \cdot \frac{3^n}{3} & =108 \end{aligned} $ $ \begin{aligned} & n \cdot 3^n=3 	imes 108 \ & n \cdot 3^n=3 	imes\left(3^3 	imes 4ight) \ & n \cdot 3^n=4 	imes 3^4 \end{aligned} $ $	herefore$ On comparison $n=4$ Hence, option (4) is correct

If $\lim _{x \rightarrow 3}\left(\frac{x^n-3^n}{x-3}\right)=108$ and $n \in \mathbf{N}$, then the value of '…

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