Let be continuous at

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

Problem

Let gg be continuous at x=2x=2, and define h(x)=sin⁡(g(x))h(x)=\sin\bigl(g(x)\bigr). Which statement best justifies the value of lim⁡x→2h(x)\displaystyle \lim_{x\to 2} h(x)? (a) Because gg is continuous at 2, lim⁡x→2g(x)=0\displaystyle\lim_{x\to 2} g(x)=0, so lim⁡x→2h(x)=0.\displaystyle\lim_{x\to 2} h(x)=0. (b) The continuity of gg guarantees g(2)=2g(2)=2, hence lim⁡x→2h(x)=sin⁡2.\displaystyle \lim_{x\to 2} h(x)=\sin 2. (c) Since gg is continuous at 2 and sin⁡x\sin x is continuous for all real xx, the composition hh is continuous at 2, giving lim⁡x→2h(x)=sin⁡(g(2)).\displaystyle \lim_{x\to 2} h(x)=\sin\bigl(g(2)\bigr). (d) Because sin⁡x\sin x oscillates infinitely often, the limit of hh does not exist no matter what gg is.

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