A function f(x) is defined by f(x) = \begin{cases} \frac{1 - \cos(6x)}{x \tan(3x)}, & x \neq 0 \\ p + 4, & x = 0 \end{cases} If f(x) is continuous at x = 0, what is the value of the constant p?

A function f(x) is defined by f(x) = \begin{cases} \frac{1 - \cos(6x)}{x \tan(3x)}, & x \n… — JAMB UTME

A function

f(x)

is defined by

f(x) = \begin{cases} \frac{1 - \cos(6x)}{x \tan(3x)}, & x \neq 0 \\ p + 4, & x = 0 \end{cases}

f(x)

is continuous at

x = 0

, what is the value of the constant

p

The value of the constant

p

2

For the piecewise function to be continuous at

x = 0

, the limit

\lim_{x \to 0} f(x)

must exist and equal

f(0) = p + 4

. By applying the double-angle identity

1 - \cos(6x) = 2\sin^2(3x)

and using the standard limit

\lim_{u \to 0} \frac{\sin u}{u} = 1

, the limit evaluates to

6

. Equating

p + 4 = 6

yields

p = 2

State the continuity condition at

x = 0

\lim_{x \to 0} f(x) = f(0) = p + 4

For a function to be continuous at a point, its limit at that point must equal the defined function value.

Apply the trigonometric identity

1 - \cos(6x) = 2\sin^2(3x)

\frac{1 - \cos(6x)}{x \tan(3x)} = \frac{2\sin^2(3x)}{x \cdot \frac{\sin(3x)}{\cos(3x)}} = 2\cos(3x) \cdot \frac{\sin(3x)}{x}

Rewriting

\tan(3x)

\frac{\sin(3x)}{\cos(3x)}

allows cancellation of one

\sin(3x)

factor.

Evaluate the limit as

x \to 0

\lim_{x \to 0} \left[2\cos(3x) \cdot 3 \cdot \frac{\sin(3x)}{3x}\right] = 2(1)(3)(1) = 6

Using the standard trigonometric limit \lim_{u \to 0} \frac{\sin(u)}{u} = 1 and \cos(0) = 1.

Equate the limit to

f(0)

and solve for

p

p + 4 = 6 \implies p = 2

Subtracting 4 from both sides isolates the parameter

p

Continuity of a Piecewise Function using Trigonometric Limits

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