Question

Evaluate the limit:$\underset{x\to 0}{Lim}\frac{\sqrt{4+x}-\sqrt[3]{38+x}}{x}=$

• A. $\frac{-1}{6}$
• B. $\frac{1}{6}$
• C. $\frac{1}{12}$
• D. $\frac{-1}{12}$

Answer 1

Answer: (B)

$\frac{1}{6}$

${lim}_{x\to 0}\frac{{(4+x)}^{\frac{1}{2}}-{(8+x)}^{\frac{1}{3}}}{x}$

Using L-H-R.

$\Rightarrow {lim}_{x\to 0}\frac{1}{2{(4+x)}^{\frac{1}{2}}}-\frac{1}{3{(8+x)}^{\frac{2}{3}}}\Rightarrow \frac{1}{2\sqrt{4}}-\frac{1}{3\sqrt[3]{38^2}}\Rightarrow \frac{1}{4}-\frac{1}{12}\Rightarrow \frac{1}{6}$