A student wishes to compute and first rewrites as

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

Problem

A student wishes to compute lim⁡x→0tan⁡xx\displaystyle\lim_{x\to 0}\frac{\tan x}{x} and first rewrites tan⁡x\tan x as sin⁡xcos⁡x\dfrac{\sin x}{\cos x}. Which of the following best explains why this rewrite can simplify the evaluation of the limit? (a) Because after rewriting, the xx in the denominator cancels with the xx inside sin⁡x\sin x, leaving 1cos⁡x\dfrac{1}{\cos x}, whose limit as x→0x\to 0 is 00. (b) Because expressing tan⁡x\tan x as sin⁡xcos⁡x\dfrac{\sin x}{\cos x} turns the limit into a 0/00/0 indeterminate form to which L’Hospital’s Rule can be applied, giving sec⁡20=1\sec^{2}0=1. (c) Because sin⁡xx\dfrac{\sin x}{x} and cos⁡x\cos x each have well-known limits as x→0x\to 0, enabling the expression to be factored into familiar pieces. (d) Because sin⁡xcos⁡x\dfrac{\sin x}{\cos x} is continuous at x=0x=0, so the entire expression is continuous there and the limit equals its value at x=0x=0.

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