Question

Find $n$, if ${lim}_{x\to 2}\frac{x^n-2^n}{x-2}=80$, $n$ being a positive integer.

Answer 1

Given, ${lim}_{x\to 2}\frac{x^n-2^n}{x-2}=80$

This is $\frac{0}{0}$ form.

So, by using the L'Hospital rule

${lim}_{x\to 2}\frac{{nx}^{n-1}-0}{1}=80{lim}_{x\to 2}{nx}^{n-1}=80$

Thus $n{2}^{n-1}=80$

Which is true for $n=5$

So, $n=5$