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If $l=\lim _{x \rightarrow 0} \frac{x}{|x|+x^2}$, then the value of $l$ is
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Verified Answer
The correct answer is:
non-existant
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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