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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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Verified Answer
The correct answer is:
4
$\lim _{x \rightarrow 3} \frac{x^n-3^n}{x-3}=108$
LHS is in $\div$ from, so apply L-Hospital rule
$$
\begin{aligned}
\lim _{x \rightarrow 3} \frac{n \cdot x^{n-1}-0}{1-0} & =108 \\
\lim _{x \rightarrow 3} n \cdot x^{n-1} & =108 \\
n \cdot 3^{n-1} & =108 \\
n \cdot \frac{3^n}{3} & =108
\end{aligned}
$$
$$
\begin{aligned}
& n \cdot 3^n=3 \times 108 \\
& n \cdot 3^n=3 \times\left(3^3 \times 4\right) \\
& n \cdot 3^n=4 \times 3^4
\end{aligned}
$$
$\therefore$ On comparison $n=4$
Hence, option (4) is correct.
LHS is in $\div$ from, so apply L-Hospital rule
$$
\begin{aligned}
\lim _{x \rightarrow 3} \frac{n \cdot x^{n-1}-0}{1-0} & =108 \\
\lim _{x \rightarrow 3} n \cdot x^{n-1} & =108 \\
n \cdot 3^{n-1} & =108 \\
n \cdot \frac{3^n}{3} & =108
\end{aligned}
$$
$$
\begin{aligned}
& n \cdot 3^n=3 \times 108 \\
& n \cdot 3^n=3 \times\left(3^3 \times 4\right) \\
& n \cdot 3^n=4 \times 3^4
\end{aligned}
$$
$\therefore$ On comparison $n=4$
Hence, option (4) is correct.
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