AP EAMCET · Maths · Limits
\(\lim _{x \rightarrow 2} \frac{\sqrt{1+4 x}-\sqrt{3+3 x}}{x^3-8}=\)
- A \(\frac{1}{72}\)
- B \(\frac{1}{36}\)
- C \(\frac{1}{24}\)
- D \(\frac{1}{12}\)
Answer & Solution
Correct Answer
(A) \(\frac{1}{72}\)
Step-by-step Solution
Detailed explanation
\(\lim _{x \rightarrow 2} \frac{\sqrt{1+4 x}-\sqrt{3+3 x}}{x^3-8}\left(\frac{0}{0}\right.\) form \()\) \(=\lim _{x \rightarrow 2} \frac{\frac{4}{2 \sqrt{1+4 x}}-\frac{3}{2 \sqrt{3+3 x}}}{3 x^2}=\frac{\frac{4}{6}-\frac{3}{6}}{12}=\frac{1}{72}\)
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