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MHT CET · Maths · Limits

If \(l=\lim _{x \rightarrow 0} \frac{x}{|x|+x^2}\), then the value of \(l\) is

  1. A \(1\)
  2. B \(-1\)
  3. C \(2\)
  4. D non-existant
Verified Solution

Answer & Solution

Correct Answer

(D) non-existant

Step-by-step Solution

Detailed explanation

Let \(\mathrm{f}(x)=\frac{x}{|x|+x^2}\)
\(\begin{aligned}
& \lim _{x \rightarrow 0^{-}} \mathrm{f}(x)=\lim _{x \rightarrow 0} \frac{x}{-x+x^2}=\lim _{x \rightarrow 0} \frac{1}{-1+x}=-1 \\
& \lim _{x \rightarrow 0^{+}} \mathrm{f}(x)=\lim _{x \rightarrow 0} \frac{x}{x+x^2}=\lim _{x \rightarrow 0} \frac{1}{1+x}=1
\end{aligned}\)
Here, \(\lim _{x \rightarrow 0^{-}} \mathrm{f}(x) \neq \lim _{x \rightarrow 0^{+}} \mathrm{f}(x)\)
\(\therefore \quad\) Value of \(l\) is non-existant