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Question: Answered & Verified by Expert
$$
\lim _{x \rightarrow 2} \frac{\sqrt[3]{6+x}-\sqrt[3]{10-x}}{x-2}=
$$
MathematicsLimitsTS EAMCETTS EAMCET 2023 (13 May Shift 2)
Options:
  • A $\frac{1}{8}$
  • B $\frac{1}{4}$
  • C $\frac{1}{2}$
  • D $\frac{1}{16}$
Solution:
1032 Upvotes Verified Answer
The correct answer is: $\frac{1}{4}$
$\begin{aligned} & \lim _{x \rightarrow 2} \frac{\sqrt[3]{6+x}-\sqrt[3]{10-x}}{x-2} \\ = & \lim _{x \rightarrow 2} \frac{(\sqrt[3]{6+x}-\sqrt[3]{10-x})\left((6+x)^{2 / 3}+(6+x)^{1 / 3}(10-x)^{1 / 3}+(10-x)^{2 / 3}\right)}{(x-2)\left((6+x)^{2 / 3}+(6+x)^{1 / 3}(10-x)^{1 / 3}+(10-x)^{2 / 3}\right)} \\ = & \lim _{x \rightarrow 2} \frac{(6+x)-(10-x)}{(x-2)\left(\sqrt[3]{(6+x)^2}+\sqrt[3]{(6+x)(10-x)}+\sqrt[3]{(10-x)^2}\right.} \\ = & \lim _{x \rightarrow 2} \frac{2(x-2)}{(x-2)\left(\sqrt[3]{(6+x)^2}+\sqrt[3]{(6+x)(10-x)}+\sqrt[3]{(10-x)^2}\right.} \\ = & \frac{2}{4+4+4}=\frac{1}{6}\end{aligned}$

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