Question

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

Answer 1

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

${\sum }_{k=1}^{100}{x}^k=x^1+x^2+...........+x^{100}=\frac{x(x^{100}-1)}{x-1}=\frac{x^{101}-x}{x-1}$

${lim}_{x\to 1}\frac{\frac{x^{101}-x}{x-1}-100}{x-1}$

$lim\frac{x^{101}-x-100x+100}{(x-1{)}^2}$

It is in $\frac{0}{0}$ form,$\therefore$ we apply L-Hospital rule,

${lim}_{x\to 1}\frac{101x^{100}-1-100}{2x-2}$

We apply L-Hospital rule again,

${lim}_{x\to 1}\frac{101\times 100x^{99}}{2}=\frac{101\times 100}{2}=\frac{10100}{2}=5050$