Question

Evaluate the following limit:
$\underset{x\to 0}{\text{lim}}\frac{(x+1{)}^5-1}{x}$

Answer 1

Here $\underset{x\to 0}{\text{lim}}\frac{(x+1{)}^5-1}{x}\left[\frac{0}{0}\text{form}\right]$

= $\underset{x\to 0}{\text{lim}}\frac{(x+1{)}^5-1}{(x+1)-1}$

Putting x+1 = y, as x $\to$ 0, y $\to$ 1

$\therefore \underset{y\to 0}{\text{lim}}\frac{y^5-1}{y-1}=5.(1{)}^{5-1}=5\times 1$ = 5.

$[\because \underset{x\to a}{\text{lim}}\frac{x^n-a^n}{x-a}=n.a^{n-1}]$