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Evaluate a 0/0 limit using L'Hopital's Rule

Consider the function g(x)=sin⁡(x)xg(x) = \dfrac{\sin(x)}{x}.

Evaluate the limit lim⁡x→0sin⁡(x)x\lim_{x \to 0} \frac{\sin(x)}{x} using L'Hopital's Rule. Be sure to verify that the limit is in an indeterminate form before applying the rule, and show the derivative of the numerator and denominator separately.

Hint

First check what happens if you plug x=0x = 0 directly into both the numerator and denominator. If you get 0/00/0, differentiate the numerator and denominator separately (not using the quotient rule), then substitute again.

Step-by-step solution

Substituting x=0x=0 directly gives sin⁡(0)/0=0/0\sin(0)/0 = 0/0, which is an indeterminate form, so L'Hopital's Rule applies. Differentiate the numerator and denominator separately: the derivative of sin⁡(x)\sin(x) is cos⁡(x)\cos(x), and the derivative of xx is 11.

So the limit becomes lim⁡x→0cos⁡(x)1=cos⁡(0)=1.\lim_{x\to 0} \frac{\cos(x)}{1} = \cos(0) = 1. Therefore lim⁡x→0sin⁡(x)x=1\lim_{x\to 0} \dfrac{\sin(x)}{x} = 1.

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