AP EAMCET · Maths · Continuity and Differentiability
If the function defined by \(f(x)=\frac{\log (1+x)^{1+x}}{x^2}-\frac{1}{x}, x \neq 0\) is continuous at \(x=0\), then \(6 f(0)\) is equal to
- A 2
- B 3
- C 1
- D 6
Answer & Solution
Correct Answer
(B) 3
Step-by-step Solution
Detailed explanation
We have, \(f(x)=\frac{\log (1+x)^{1+x}}{x^2}-\frac{1}{x}\) \[ \begin{aligned} \Rightarrow \quad f(x) & =\frac{(1+x) \log (1+x)}{x^2}-\frac{1}{x} \\ & =\frac{(1+x) \log (1+x)-x}{x^2} \end{aligned} \] \(f(x)\) is continuous at \(x=0\)…
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