TS EAMCET · Maths · Limits
For \(a \neq 0\) and \(b \neq 0\), if the real valued function \(f(x)=\frac{\sqrt[5]{a(625+x)}-5}{\sqrt[4]{625+b x}-5}\) is continuous at \(x=0\), then \(f(0)=\)
- A \(\frac{4 b}{5}\)
- B \(\frac{5 b}{4}\)
- C \(\frac{5}{4 b}\)
- D \(\frac{4}{5 b}\)
Answer & Solution
Correct Answer
(D) \(\frac{4}{5 b}\)
Step-by-step Solution
Detailed explanation
For continuity at \(x=0\), the numerator must be 0: \(\sqrt[5]{a(625)}-5 = 0 \implies 625a = 5^5 \implies a = 5\) Using L'Hôpital's Rule for \(f(0) = \lim_{x \to 0} \frac{\frac{d}{dx}(\sqrt[5]{a(625+x)}-5)}{\frac{d}{dx}(\sqrt[4]{625+b x}-5)}\): Numerator derivative at \(x=0\):…
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