Question

$\underset{x\to 0}{lim}\frac{{(1+x)}^5-1}{{(1+x)}^3-1}=$

• A. $0$
• B. $1$
• C. $\frac{5}{3}$
• D. $\frac{3}{5}$

Answer 1

The correct option is C $\frac{5}{3}$

We have,

${lim}_{x\to 0}\frac{(1+x{)}^5-1}{(1+x{)}^3-1}$

This is the $\frac{0}{0}$ form.

Apply L-hospital rule,

${lim}_{x\to 0}\frac{5(1+x{)}^4-0}{3(1+x{)}^2-0}$

${lim}_{x\to 0}\frac{5(1+x{)}^4}{3(1+x{)}^2}$

${lim}_{x\to 0}\frac{5(1+x{)}^2}{3}$

$=\frac{5(1+0{)}^2}{3}$

$=\frac{5}{3}$

Hence, this is the answer.