MHT CET · Maths · Limits
If \(l=\lim _{x \rightarrow 0} \frac{x}{|x|+x^2}\), then the value of \(l\) is
- A \(1\)
- B \(-1\)
- C \(2\)
- D non-existant
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
\(\begin{aligned}
& \lim _{x \rightarrow 0^{-}} \mathrm{f}(x)=\lim _{x \rightarrow 0} \frac{x}{-x+x^2}=\lim _{x \rightarrow 0} \frac{1}{-1+x}=-1 \\
& \lim _{x \rightarrow 0^{+}} \mathrm{f}(x)=\lim _{x \rightarrow 0} \frac{x}{x+x^2}=\lim _{x \rightarrow 0} \frac{1}{1+x}=1
\end{aligned}\)
Here, \(\lim _{x \rightarrow 0^{-}} \mathrm{f}(x) \neq \lim _{x \rightarrow 0^{+}} \mathrm{f}(x)\)
\(\therefore \quad\) Value of \(l\) is non-existant
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