Question

Evaluate: $\underset{x\to 0}{lim}\frac{(1+x{)}^3-1-3x}{(1+x{)}^2-1-2x}=$

• A. $2$
• B. $3$
• C. $5$
• D. $-2$

Answer 1

Answer: (B)

$3$

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

It is of the form $\frac{0}{0}$, so applying L-Hospital's rule

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

Again of the form, $\frac{0}{0}$, so applying L-Hospital's rule

$=\underset{x\to 0}{lim}\frac{6(x+1)}{2}$

$=3$

Hence option ${}^{\prime }{B}^{\prime }$ is the answer.