If l = lim_{x rightarrow 0} frac{x}{|x| + x^2 | vedclass

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(D) Let $f(x) = \frac{x}{|x| + x^2}$.For the left-hand limit as $x \rightarrow 0^-$,we have $|x| = -x$:$\lim_{x \rightarrow 0^-} f(x) = \lim_{x \right

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non-existent

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(D) Let $f(x) = \frac{x}{|x| + x^2}$.For the left-hand limit as $x ightarrow 0^-$,we have $|x| = -x$:$\lim_{x ightarrow 0^-} f(x) = \lim_{x ightarrow 0} \frac{x}{-x + x^2} = \lim_{x ightarrow 0} \frac{1}{-1 + x} = -1$.For the right-hand limit as $x ightarrow 0^+$,we have $|x| = x$:$\lim_{x ightarrow 0^+} f(x) = \lim_{x ightarrow 0} \frac{x}{x + x^2} = \lim_{x ightarrow 0} \frac{1}{1 + x} = 1$.Since $\lim_{x ightarrow 0^-} f(x) 
eq \lim_{x ightarrow 0^+} f(x)$,the limit does not exist.

If $l = \lim_{x \rightarrow 0} \frac{x}{|x| + x^2}$,then the value of $l$ is

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