Q102 / 160 · TS EAMCET 2023 (13 May Shift 2) · Mathematics
$$
\lim _{x \rightarrow 2} \frac{\sqrt[3]{6+x}-\sqrt[3]{10-x}}{x-2}=
$$
- A$\frac{1}{8}$
- Correct answerB$\frac{1}{4}$
- C$\frac{1}{2}$
- D$\frac{1}{16}$
Solution
$\begin{aligned} & \lim _{x \rightarrow 2} \frac{\sqrt[3]{6+x}-\sqrt[3]{10-x}}{x-2} \\ = & \lim _{x \rightarrow 2} \frac{(\sqrt[3]{6+x}-\sqrt[3]{10-x})\left((6+x)^{2 / 3}+(6+x)^{1 / 3}(10-x)^{1 / 3}+(10-x)^{2 / 3}\right)}{(x-2)\left((6+x)^{2 / 3}+(6+x)^{1 / 3}(10-x)^{1 / 3}+(10-x)^{2 / 3}\right)} \\ = & \lim _{x \rightarrow 2} \frac{(6+x)-(10-x)}{(x-2)\left(\sqrt[3]{(6+x)^2}+\sqrt[3]{(6+x)(10-x)}+\sqrt[3]{(10-x)^2}\right.} \\ = & \lim _{x \rightarrow 2} \frac{2(x-2)}{(x-2)\left(\sqrt[3]{(6+x)^2}+\sqrt[3]{(6+x)(10-x)}+\sqrt[3]{(10-x)^2}\right.} \\ = & \frac{2}{4+4+4}=\frac{1}{6}\end{aligned}$
Hence, the correct answer is option: 'B'.