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MHT CET · Maths · Limits

If \(\mathrm{f}(x)=\frac{(27-2 x)^{\frac{1}{3}}-3}{9-3(243+5 x)^{\frac{1}{5}}}, x \neq 0\) is continuous at \(x=0\), then the value of \(f(0)\) is

  1. A \(\frac{2}{3}\)
  2. B \(6\)
  3. C \(2\)
  4. D \(\frac{1}{3}\)
Verified Solution

Answer & Solution

Correct Answer

(C) \(2\)

Step-by-step Solution

Detailed explanation

\(\mathrm{f}(0) = \lim_{x \to 0} \mathrm{f}(x) = \lim_{x \to 0} \frac{(27-2 x)^{\frac{1}{3}}-3}{9-3(243+5 x)^{\frac{1}{5}}}\) \(\mathrm{f}(0) = \lim_{x \to 0} \frac{3\left(\left(1-\frac{2x}{27}\right)^{\frac{1}{3}}-1\right)}{-9\left(\left(1+\frac{5x}{243}\right)^{\frac{1}{5}}-1\right)}\)