Question:
Find ' $n$ ', if $\lim _{x \rightarrow 2} \frac{x^{n}-2^{n}}{x-2}=80, n \in \mathbf{N}$
Solution:
Given $\lim _{x \rightarrow 2} \frac{x^{n}-2^{n}}{x-2}=80$
We know that $\lim _{x \rightarrow a} \frac{x^{n^{n}}-a^{n}}{x-a}=n a^{n-1}$
By using this formula we get
$\Rightarrow \mathrm{n} 2^{\mathrm{n}-1}=80=5 \times 2^{4}=5 \times 2^{5-1}$
$\Rightarrow \mathrm{n}=5$
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