Question

$li{m}_{x\to 0}\frac{(1+x{)}^4-1}{(1+x{)}^3-1}=..........$

• A. $\frac{4x}{3}$
• B. $4x^3$
• C. $\frac{4}{3}$
• D. $\frac{3x}{4}$

Answer 1

The correct option is C $\frac{4}{3}$

$let(1+x)=twhenx\to 0,t\to 1{lim}_{t\to 1}\frac{t^4-1}{t^3-1}\Rightarrow \frac{(t-1)(t+1)(t^2+1)}{(t-1)(t^2+t+1)}\Rightarrow \frac{(t+1)(t^2+1)}{(t^2+t+1)}\Rightarrow \frac{2\times 2}{3}\Rightarrow \frac{4}{3}$