Question

Evaluate: $\underset{x\to 1}{lim}\frac{\left[\sum _{K=1}^{100}{x}^k\right]-100}{x-1}$

Answer 1

$\underset{x\to 1}{lim}\frac{\left(\sum _{k=1}^{100}{x}^k\right)-100}{x-1}$

$=\underset{x\to 1}{lim}\frac{\left(\sum _{k=0}^{100}{x}^k\right)-101}{x-1}$

$=\underset{x\to 1}{lim}\frac{\left(\frac{x^{101}-1}{(x-1)}\right)-101}{x-1}$

$=\underset{x\to 1}{lim}\frac{(x^{101}-1)-101(x-1)}{(x-1{)}^2}$ [$\frac{0}{0}$ form]

Using L'Hospital's rule we get,

$=\underset{x\to 1}{lim}\frac{\left(101x^{100}\right)-101}{2(x-1)}$ [$\frac{0}{0}$ form]

Using L'Hospital's rule we get,

$=\underset{x\to 1}{lim}\frac{\left(101.100x^{99}\right)}{2}$

$=101\times 50=5050$.