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
- A \(\frac{2}{3}\)
- B \(6\)
- C \(2\)
- D \(\frac{1}{3}\)
Answer & Solution
Correct Answer
(C) \(2\)
Step-by-step Solution
Detailed explanation
\(\mathrm{f}(0) = \lim_{x \to 0} \mathrm{f}(x) = \lim_{x \to 0} \frac{(27-2 x)^{\frac{1}{3}}-3}{9-3(243+5 x)^{\frac{1}{5}}}\) \(\mathrm{f}(0) = \lim_{x \to 0} \frac{3\left(\left(1-\frac{2x}{27}\right)^{\frac{1}{3}}-1\right)}{-9\left(\left(1+\frac{5x}{243}\right)^{\frac{1}{5}}-1\right)}\)
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