A graph reveals the following one-sided limits for functions and at : _ x 1^ -…

MATHEMATICS · Single correct · AP · Calculus AB · Limits and Continuity · Limits and Continuity

Problem

A graph reveals the following one-sided limits for functions rr and ss at x=1x=1: lim⁡x→1−r(x)=0,lim⁡x→1+r(x)=4,andlim⁡x→1s(x)=−2.\lim_{x\to 1^{-}} r(x)=0, \qquad \lim_{x\to 1^{+}} r(x)=4, \qquad \text{and} \qquad \lim_{x\to 1} s(x)=-2. Consider h(x)=r(x)s(x)h(x)=\dfrac{r(x)}{s(x)}. Which statement about lim⁡x→1h(x)\displaystyle \lim_{x\to 1} h(x) is true? (a) lim⁡x→1h(x)=0\displaystyle \lim_{x\to 1} h(x)=0 because both one–sided limits equal 00. (b) lim⁡x→1h(x)\displaystyle \lim_{x\to 1} h(x) is undefined because lim⁡x→1s(x)=0\displaystyle \lim_{x\to 1} s(x)=0. (c) lim⁡x→1−h(x)=0 and lim⁡x→1+h(x)=−2\displaystyle \lim_{x\to 1^{-}} h(x)=0\text{ and }\lim_{x\to 1^{+}} h(x)=-2; lim⁡x→1h(x)\displaystyle \lim_{x\to 1} h(x)\text{ does not exist}. (d) lim⁡x→1h(x)=−2\displaystyle \lim_{x\to 1} h(x)=-2 because the right–hand limit of rr dominates the fraction.

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