Question

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

(a) n
(b) 1
(c) –n
(d) 0

Answer 1

$$\begin{gathered} \begin{array}{c}\lim \\ x\to 0\end{array}\frac{{\left(1+x\right)}^n-1}{x} \\ \mathrm{since}{\left(1+x\right)}^n=1+nx+\frac{n\left(n-1\right)}{2}{x}^2+..................+x^n \\ \mathrm{i}.\mathrm{e}.\begin{array}{c}\lim \\ x\to 0\end{array}\frac{\left[1+nx+{\displaystyle \frac{n\left(n-1\right)x^2}{2}}+..................+x^n\right]-1}{x} \\ \mathrm{i}.\mathrm{e}.\begin{array}{c}\lim \\ x\to 0\end{array}\frac{x\left[n+{\displaystyle \frac{n\left(n-1\right)}{2}}x+..................+x^{n-1}\right]}{x} \\ \mathrm{i}.\mathrm{e}.\begin{array}{c}\lim \\ x\to 0\end{array}\left[n+\frac{n\left(n-1\right)x}{2}+..................+x^{n-1}\right] \\ \mathrm{i}.\mathrm{e}.\begin{array}{c}\lim \\ x\to 0\end{array}n+\begin{array}{c}\lim \\ x\to 0\end{array}\frac{n\left(n-1\right)x}{2}+..................+\begin{array}{c}\lim \\ x\to 0\end{array}{x}^n=n \end{gathered}$$

Hence, the correct answer is option A.