MHT CET · Maths · Limits
\(\lim _{x \rightarrow 0} \frac{2 x}{|x|+x^2}=\)
- A Limit exists
- B Limit does not exists
- C \(2\)
- D \(-2\)
Answer & Solution
Correct Answer
(B) Limit does not exists
Step-by-step Solution
Detailed explanation
\(\lim _{x \rightarrow 0} \frac{2 x}{|x|+x^2}\)
L.H.L \(\lim _{x \rightarrow 0^{-}} \frac{2 x}{-x+x^2}=\lim _{x \rightarrow 0^{-}} \frac{2}{-1+x}=-\infty\)
R.H.L \(\lim _{x \rightarrow 0^{+}} \frac{2 x}{x+x^2}=\lim _{x \rightarrow 0^{+}} \frac{2}{1+x}=\infty\)
\(\because\) L.H.L \(\neq\) R.H.L \(\Rightarrow\) limit does not exist
L.H.L \(\lim _{x \rightarrow 0^{-}} \frac{2 x}{-x+x^2}=\lim _{x \rightarrow 0^{-}} \frac{2}{-1+x}=-\infty\)
R.H.L \(\lim _{x \rightarrow 0^{+}} \frac{2 x}{x+x^2}=\lim _{x \rightarrow 0^{+}} \frac{2}{1+x}=\infty\)
\(\because\) L.H.L \(\neq\) R.H.L \(\Rightarrow\) limit does not exist
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