Question

Evaluate: $\underset{x\to 0}{lim}\frac{\sqrt{4+x}-\sqrt[3]{38+3x}}{x}$

• A. $-\frac{1}{2}$
• B. $\frac{1}{2}$
• C. $-3$
• D. $0$

Answer 1

Answer: (D)

$0$

Given: $\underset{x\to 0}{lim}\frac{\sqrt{4+x}-\sqrt[3]{38+3x}}{x}$

Here, as $x\to 0$ the expression becomes $\frac{0}{0}$ form.

Hence we can use L'Hospital's rule.

If we want to evaluate $\underset{x\to a}{lim}\frac{f\left(x\right)}{g\left(x\right)}$ and it is of indeterminate form $\left(\frac{0}{0}or\frac{\infty }{\infty }\right)$

we can evaluate the expression using

$\underset{x\to a}{lim}\frac{f\left(x\right)}{g\left(x\right)}=\underset{x\to a}{lim}\frac{f^{\prime }\left(x\right)}{g^{\prime }\left(x\right)}$

Here, $f\left(x\right)=\sqrt{4+x}-\sqrt[3]{38+3x}$ and $g\left(x\right)=x$.

$f^{\prime }\left(x\right)=\frac{1}{2(4+x{)}^{1/2}}-\frac{1\times 3}{3(8+3x{)}^{2/3}}$ ......$(Using\frac{d}{dx}{x}^n=n{x}^{n-1})$

and $g^{\prime }\left(x\right)=1$

$\underset{x\to a}{lim}\frac{f^{\prime }\left(x\right)}{g^{\prime }\left(x\right)}=\underset{x\to 0}{lim}\frac{\frac{1}{2(4+x{)}^{1/2}}-\frac{3}{3(8+3x{)}^{2/3}}}{1}$

$=\frac{1}{2\sqrt{4}}-\frac{1}{8^{2/3}}$

$=\frac{1}{2\times 2}-\frac{1}{(2^3{)}^{2/3}}$

$=\frac{1}{2\times 2}-\frac{1}{4}$

$=\frac{1}{4}-\frac{1}{4}$

Answer is $0$