Question

What is $\underset{x\to 0}{lim}\frac{(1+x{)}^n-1}{x}$ equal to ?

• A. 0
• B. 1
• C. n
• D. n-1

Answer 1

The correct option is C n

This can be viewed to be of form $\underset{h\to 0}{lim}\frac{f(a+h)-f\left(a\right)}{h}$, where $f\left(x\right)=x^n$ and $a=1$

We know that $\underset{h\to 0}{lim}\frac{f(a+h)-f\left(a\right)}{h}$ can be written as $f^{{}^{\prime }}\left(a\right)$ ie. derivative of $f\left(x\right)atx=a$

Hence, the given limit reduces to $\frac{d\left(x^n\right)}{dx}forx=1$

$f\left(x\right)=x^n$ and $f\left(1\right)=1$

$\Rightarrow f^{{}^{\prime }}\left(x\right)=n{x}^{(n-1)}$

$\Rightarrow f^{{}^{\prime }}\left(1\right)=n$

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