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Limit of tan(x)/x using known limit laws

Recall that tan⁡(x)=sin⁡(x)cos⁡(x)\tan(x) = \dfrac{\sin(x)}{\cos(x)}, and that lim⁡x→0sin⁡(x)x=1\displaystyle\lim_{x\to 0}\frac{\sin(x)}{x} = 1.

Evaluate the limit lim⁡x→0tan⁡(x)x.\lim_{x\to 0}\frac{\tan(x)}{x}.

Hint

Rewrite tan⁡(x)/x\tan(x)/x as a product of sin⁡(x)/x\sin(x)/x and 1/cos⁡(x)1/\cos(x), then apply the product rule for limits along with the known limits lim⁡x→0sin⁡(x)/x=1\lim_{x\to 0}\sin(x)/x = 1 and direct substitution for cos⁡(x)\cos(x).

Step-by-step solution

Rewrite the expression as tan⁡(x)x=sin⁡(x)x⋅1cos⁡(x).\frac{\tan(x)}{x} = \frac{\sin(x)}{x}\cdot\frac{1}{\cos(x)}. This splits the original function into two factors whose limits are individually known or easy to compute as x→0x \to 0.

By the product rule for limits, lim⁡x→0tan⁡(x)x=(lim⁡x→0sin⁡(x)x)(lim⁡x→01cos⁡(x)).\lim_{x\to 0}\frac{\tan(x)}{x} = \left(\lim_{x\to 0}\frac{\sin(x)}{x}\right)\left(\lim_{x\to 0}\frac{1}{\cos(x)}\right). The first factor equals 11 by the standard trigonometric limit, and the second factor equals 1/cos⁡(0)=11/\cos(0) = 1 by direct substitution since cos⁡(x)\cos(x) is continuous at 00. Multiplying these gives 1⋅1=11 \cdot 1 = 1.

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