Question

Let $\underset{x\to 1}{lim}\frac{x^a-ax+a-1}{(x-1{)}^2}=f\left(a\right)$. Then the value of $f\left(4\right)$ is ?

• A. $2$
• B. $4$
• C. $6$
• D. $8$

Answer 1

The correct option is C $6$

Put $x=1+h$. Then,
$f\left(a\right)=\underset{h\to 0}{lim}\frac{(1+h{)}^a-a(1+h)+a-1}{h^2}$

$=\underset{h\to 0}{lim}\frac{(1+ah+\frac{a(a-1)}{2!}{h}^2+....)-a-ah+a-1}{h^2}$

$\therefore f\left(a\right)=\frac{a(a-1)}{2}$

$\therefore f\left(4\right)=6$
Hence, option 'C' is correct.