MHT CET · Maths · Limits
\(\lim _{x \rightarrow 0} \frac{|x|}{|x|+x^2}=\)
- A \(0\)
- B \(1\)
- C \(-1\)
- D \(\frac{1}{2}\)
Answer & Solution
Correct Answer
(B) \(1\)
Step-by-step Solution
Detailed explanation
For \(x \rightarrow 0^+\): \(\lim _{x \rightarrow 0^+} \frac{x}{x+x^2} = \lim _{x \rightarrow 0^+} \frac{1}{1+x} = 1\). For \(x \rightarrow 0^-\): \(\lim _{x \rightarrow 0^-} \frac{-x}{-x+x^2} = \lim _{x \rightarrow 0^-} \frac{1}{1-x} = 1\).
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