Question

$\underset{x\to 0}{lim}\frac{\sqrt[3]{31+3x}-1-x}{(1+x{)}^{101}-1-101x}$ has the value equal to -

• A. $-\frac{3}{5050}$
• B. $-\frac{1}{5050}$
• C. $\frac{1}{5051}$
• D. $\frac{1}{4950}$

Answer 1

Answer: (B)

$-\frac{1}{5050}$

$\underset{x\to 0}{lim}\frac{(1+3x{)}^{1/3}-1-x}{(1+x{)}^{101}-1-101x}$
$=\underset{x\to 0}{lim}\frac{\frac{3}{3}(1+3x{)}^{-2/3}-1}{\left(101\right)(1+x{)}^{100}-101}$(By L' Hospital's rule)
$=\underset{x\to 0}{lim}\frac{-\frac{2}{3}(1+3x{)}^{-5/3}.3}{\left(101\right)\left(100\right)(1+x{)}^{99}}=-\frac{1}{5050}$
Hence, option 'B' is correct.