WEBVTT

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Here is the problem: (1-cos(3x))/x squared.

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Find the limit of (1 minus cos(3x)) / x squared as x → 0.

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Indeterminate form 0 over 0 — apply L prime Hôpital prime s rule. Differentiate the numerator and the denominator separately (this is not the quotient rule).

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Indeterminate form 0 over 0 — apply L prime Hôpital prime s rule. Differentiate the numerator and the denominator separately (this is not the quotient rule).

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Direct substitution. The expression is continuous at x equals 0, so the limit equals its value there.

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So the answer is Limit equals 9 over 2 approximately 4.5.

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A limit describes the value a function approaches, not necessarily the value it takes. When substitution gives an indeterminate form, the expression is rewritten (L prime Hôpital prime s rule, a conjugate, or logarithms) until substitution works.

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Numerical values from the right agree: f(0.01) equals 4.4996625, f(0.001) equals 4.4999966, f(0.0001) equals 4.5, f(0.00001) equals 4.5000004, which checks out.

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Numerical values from the left agree: f(negative 0.01) equals 4.4996625, f(negative 0.001) equals 4.4999966, f(negative 0.0001) equals 4.5, f(negative 0.00001) equals 4.5000004, which checks out.