x→∞lim(3−x2x+2x2−x3)Divide by highest denominator power: x23−1x1+2−x=x→∞lim(x23−1x1+2−x)x→alim[g(x)f(x)]=limx→agg(x)limx→aff(x),x→aglim(x)=0With the exception of indeterminate form=limx→∞(x23−1)limx→∞(x1+2−x)x→∞lim(x1+2−x)=−∞x→∞lim(x23−1)=−1=−1−∞=∞
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