1.If l=limx→0x∣x∣+x2\displaystyle l=\lim_{x\to0}\frac{x}{|x|+x^2}l=x→0lim∣x∣+x2x, then the value of lll isa.111b.−1-1−1c.222d.non-existentLogin to continueOnly logged in users canattempt or see the solution.
