Question

Let $\underset{x\to 1}{lim}\frac{x^a-ax+a-1}{{(x-1)}^2}=f\left(a\right)$. Then the value of $f\left(101\right)$ is equal to?

• A. $5050$
• B. $5151$
• C. $4950$
• D. $101$

Answer 1

Answer: (A)

$5050$

$f\left(101\right)={lim}_{x\to 1}\frac{x^{101}-101x+101-1}{(x-1{)}^2}$
If we substitute $x=1$, we get $\frac{0}{0}$ indeterminate form.
We apply L'Hospitals Rule,
=${lim}_{x\to 1}\frac{101x^{100}-101}{2(x-1)}$
again it is $\frac{0}{0}$ indeterminate form. so applying L'Hospitals Rule,
=${lim}_{x\to 1}\frac{101.100x^{99}}{2}$
Substituting $x=1$,
=$101.50=5050$