Question

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

• A. $\frac{3}{4}$
• B. $\frac{9}{16}$
• C. $\frac{4}{3}$
• D. None of these

Answer 1

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

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

The above is in $\frac{0}{0}$ indeterminate form,so we can employ L'Hospital's Rule

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

Now we can just susbstitute the milit to obtain

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