Question

The value of $\underset{x\to 1}{lim}\frac{x^{p+1}-\left(P+1\right)x+p}{{\left(x-1\right)}^2}$ is :

• A. $P$
• B. $P-1$
• C. $P(P+1)$
• D. $\frac{P\left(P+1\right)}{2}$

Answer 1

The correct option is D $\frac{P\left(P+1\right)}{2}$

Given function $\underset{x\to 1}{lim}\frac{x^{p+1}-\left(P+1\right)x+p}{{\left(x-1\right)}^2}$ is of $\frac{0}{0}$ form at $x=1$.

So using L'Hopital's rule twice( because denominator has power $2$), we get:

$\underset{x\to 1}{lim}\frac{x^{p+1}-\left(P+1\right)x+p}{{\left(x-1\right)}^2}=\underset{x\to 1}{lim}\frac{(p+1)x^{\left(p\right)}}{2(x-1)}$

$=\underset{x\to 1}{lim}\frac{(p+1)p{x}^{(p-1)}}{2}=\frac{(p+1)p}{2}$