Question

The value of
$\underset{x\to 4}{lim}\frac{(2x{)}^{1/3}-2}{2x-8}$ is

• A. $\frac{1}{3}$
• B. $\frac{1}{6}$
• C. $\frac{1}{24}$
• D. $\frac{1}{12}$

Answer 1

The correct option is D $\frac{1}{12}$

2x - 8 = 2h (say)
$\Rightarrow$ 2x = 8 + 2h
$\Rightarrow$ x = 4 + h
$\therefore \underset{h\to 0}{lim}\frac{(8+2h{)}^{1/5}-2}{2h}$

Above form is $\left(\frac{0}{0}\right)$ by putting the value h = 0

Applting L' Hospital rule $\underset{h\to 0}{lim}\frac{\frac{1}{3}(8+2h{)}^{(\frac{1}{3}-1)}\times 2}{2}$
$=\frac{1}{3}(8{)}^{-2/3}=\frac{1}{12}$