Question

The value of $\underset{x\to 0}{lim}\frac{(1-x{)}^n-1}{x}$ is

• A. $(n-1)!$
• B. $n!$
• C. $n$
• D. $-n$

Answer 1

The correct option is D $-n$

Using L-hospital's rule,

$\underset{x\to 0}{lim}\frac{(1-x{)}^n-1}{x}=\underset{x\to 0}{lim}\frac{n(1-x{)}^{n-1}.(-1)-0}{1}$

$={lim}_{x\to 0}-n(1-x{)}^{n-1}$

$=-n(1-0{)}^{n-1}=-n$

Hence, the option 'D' is correct.