Question

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

Answer 1

We have,Evaluate $\underset{x\to 0}{lim}\frac{(x+1{)}^5-1}{x}$

At x=0,the value of the expression is $\frac{0}{0}$ form.

Applying L'Hospital Rule

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

$=\underset{x\to 0}{lim}\frac{5(x+1{)}^4\times 1-0}{1}$

$=\underset{x\to 0}{lim}5(x+1{)}^4$

Putting x=0

=$5(1{)}^4=5$