lim _{x rightarrow 0} frac{|x|}{|x|+x^2} = | vedclass

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(B) To find the limit $\lim _{x \rightarrow 0} \frac{|x|}{|x|+x^2}$,we evaluate the left-hand limit and the right-hand limit.For the left-hand limit $

Vedclass · Accepted Answer

$1$

Vedclass · Answer

(B) To find the limit $\lim _{x \rightarrow 0} \frac{|x|}{|x|+x^2}$,we evaluate the left-hand limit and the right-hand limit.For the left-hand limit $(x \rightarrow 0^-)$,we have $|x| = -x$. Thus,the expression becomes $\frac{-x}{-x+x^2} = \frac{-x}{x(x-1)} = \frac{-1}{x-1}$. As $x \rightarrow 0^-$,this approaches $\frac{-1}{0-1} = 1$.For the right-hand limit $(x \rightarrow 0^+)$,we have $|x| = x$. Thus,the expression becomes $\frac{x}{x+x^2} = \frac{x}{x(1+x)} = \frac{1}{1+x}$. As $x \rightarrow 0^+$,this approaches $\frac{1}{1+0} = 1$.Since the left-hand limit equals the right-hand limit,the limit exists and is equal to $1$.

$\lim _{x \rightarrow 0} \frac{|x|}{|x|+x^2} = $

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