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

Evaluate the limit lim⁡x→0cos⁡(x)−1xsin⁡(x)\lim_{x \to 0} \frac{\cos(x) - 1}{x\sin(x)} using L'Hopital's Rule. Be sure to verify that the limit is in an indeterminate form before applying the rule, and apply the rule again if necessary.

Hint

First check that direct substitution gives 0/00/0. Then differentiate the numerator and denominator separately (do not use the quotient rule) to form a new limit. If the new limit is still indeterminate, check the form again and apply L'Hopital's Rule a second time.

Step-by-step solution

Substituting x=0x=0 gives cos⁡(0)−10⋅sin⁡(0)=00\frac{\cos(0)-1}{0\cdot\sin(0)} = \frac{0}{0}, confirming the indeterminate form, so L'Hopital's Rule applies. Differentiating numerator and denominator separately: the derivative of cos⁡(x)−1\cos(x)-1 is −sin⁡(x)-\sin(x), and the derivative of xsin⁡(x)x\sin(x) (using the product rule) is sin⁡(x)+xcos⁡(x)\sin(x) + x\cos(x). This gives the new limit lim⁡x→0−sin⁡(x)sin⁡(x)+xcos⁡(x).\lim_{x\to 0} \frac{-\sin(x)}{\sin(x)+x\cos(x)}.

Substituting x=0x=0 again gives −sin⁡(0)sin⁡(0)+0⋅cos⁡(0)=00\frac{-\sin(0)}{\sin(0)+0\cdot\cos(0)} = \frac{0}{0}, so the form is still indeterminate and L'Hopital's Rule can be applied a second time. Differentiating again: the derivative of −sin⁡(x)-\sin(x) is −cos⁡(x)-\cos(x), and the derivative of sin⁡(x)+xcos⁡(x)\sin(x)+x\cos(x) is cos⁡(x)+cos⁡(x)−xsin⁡(x)=2cos⁡(x)−xsin⁡(x)\cos(x) + \cos(x) - x\sin(x) = 2\cos(x) - x\sin(x). This gives lim⁡x→0−cos⁡(x)2cos⁡(x)−xsin⁡(x)=−12.\lim_{x\to 0} \frac{-\cos(x)}{2\cos(x)-x\sin(x)} = \frac{-1}{2}.

Therefore the original limit equals −12-\frac{1}{2}.

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