AP EAMCET · Maths · Limits
\(\lim _{x \rightarrow-\infty} \frac{3|x|-x}{|x|-2 x}-\lim _{x \rightarrow 0} \frac{\log \left(1+x^3\right)}{\sin ^3 x}=\)
- A \(\frac{1}{3}\)
- B \(-\frac{1}{4}\)
- C \(2\)
- D \(-\frac{5}{3}\)
Answer & Solution
Correct Answer
(A) \(\frac{1}{3}\)
Step-by-step Solution
Detailed explanation
\(\lim _{x \rightarrow-\infty} \frac{3|x|-x}{|x|-2 x}-\lim _{x \rightarrow 0} \frac{\log \left(1+x^3\right)}{\sin ^3 x}\) \(\lim _{x \rightarrow-\infty} \frac{-3 x-x}{-x-2 x}-\lim _{x \rightarrow 0} \frac{\log \left(1+x^3\right)}{x^3 \frac{\left(\sin x^3 x\right)}{x^3}}\)…
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