Math · Calculus I · Worked example

Apply L’Hôpital’s rule twice

Evaluate the limit of (1 − cos x)/x² as x → 0.

lim_{x \to 0} \frac{1 - cos x}{x^{2}}

Check the form

Substituting x = 0 gives (1 − 1)/0 = 0/0, an indeterminate form, so L’Hôpital’s rule applies.

Differentiate the top and bottom

The derivative of 1 − cos x is sin x, and the derivative of x² is 2x.

lim_{x \to 0} \frac{1 - cos x}{x^{2}} = lim_{x \to 0} \frac{sin x}{2 x}

Check the form again

At x = 0, sin x/(2x) is still 0/0, so apply the rule a second time: sin x becomes cos x and 2x becomes 2.

lim_{x \to 0} \frac{sin x}{2 x} = lim_{x \to 0} \frac{cos x}{2}

Substitute

cos 0 = 1, so the limit is 1/2. A value near 0 agrees: at x = 0.1 the quotient is 0.4996.

lim_{x \to 0} \frac{cos x}{2} = \frac{1}{2}

Result

The limit is 1/2.

lim_{x \to 0} \frac{1 - cos x}{x^{2}} = \frac{1}{2}

Your turn

Evaluate the limit of sin(3x)/x as x → 0.

Show the answer and explanation

3.

The form is 0/0. Differentiating the top gives 3 cos(3x) by the chain rule, and the bottom gives 1, so the limit is 3 cos 0 = 3.

\begin{array}{l} lim_{x \to 0} \frac{sin 3 x}{x} \\ lim_{x \to 0} \frac{3 cos 3 x}{1} \\ 3 \end{array}

Keep exploring

Open the limit in Limits & one-sided behavior and change x² to x³: the tables grow without bound, negative from the left and positive from the right, so the limit does not exist.

Return to the concept →

Sources and scope

Authored study material. Tool results depend on the stated inputs and model assumptions.

OpenStax Calculus Volume 1 — L’Hôpital’s rule

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Try in the workspace

Open the example inputs, change a value and keep a useful result on your board.

Check the limit in Limits & one-sided behavior Check each step in Math Open worked example on a board Limit of a quotient in Math Reference

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