If $l=\lim _{x \rightarrow 0} \frac{x}{|x|+x^2}$, then the value of $l$ is
If $l=\lim _{x \rightarrow 0} \frac{x}{|x|+x^2}$, then the value of $l$ is
- $1$
- $-1$
- $2$
- non-existent
Solution
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-existent
Asked in: MHT CET 2023 (11 May Shift 2)
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