Q100 / 160 · AP EAMCET 2024 (21 May Shift 1) · Mathematics

$\lim _{x \rightarrow 2} \frac{\sqrt{1+4 x}-\sqrt{3+3 x}}{x^3-8}=$

  • Correct answer
    A
    $\frac{1}{72}$
  • B
    $\frac{1}{36}$
  • C
    $\frac{1}{24}$
  • D
    $\frac{1}{12}$

Solution

$\lim _{x \rightarrow 2} \frac{\sqrt{1+4 x}-\sqrt{3+3 x}}{x^3-8}\left(\frac{0}{0}\right.$ form $)$ $=\lim _{x \rightarrow 2} \frac{\frac{4}{2 \sqrt{1+4 x}}-\frac{3}{2 \sqrt{3+3 x}}}{3 x^2}=\frac{\frac{4}{6}-\frac{3}{6}}{12}=\frac{1}{72}$

Hence, the correct answer is option: 'A'.

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