A piecewise function f(x) is defined by f(x) = \begin{cases} \frac{\sin(3x) + \tan(5x)}{2x}, & x \neq 0 \\ a^2 - 5, & x = 0 \end{cases} If f(x) is continuous at x = 0, where a > 0, determine the numerical value of a.

A piecewise function f(x) is defined by f(x) = \begin{cases} \frac{\sin(3x) + \tan(5x)}{2x… — JAMB UTME

A piecewise function

f(x)

is defined by

f(x) = \begin{cases} \frac{\sin(3x) + \tan(5x)}{2x}, & x \neq 0 \\ a^2 - 5, & x = 0 \end{cases}

f(x)

is continuous at

x = 0

, where

a > 0

, determine the numerical value of

a

Answer: 3

The numerical value of

a

is 3.

For the function to be continuous at

x = 0

, the limit as

x \to 0

must equal the function value

f(0)

. Splitting the trigonometric limit gives

\frac{3}{2} + \frac{5}{2} = 4

. Setting

a^2 - 5 = 4

yields

a^2 = 9

. Because

a > 0

, taking the positive square root gives

a = 3

Evaluate the limit of the trigonometric expression as

x

approaches 0.

\lim_{x \to 0} \frac{\sin(3x) + \tan(5x)}{2x} = 4

Using standard trigonometric limit principles: \lim_{x \to 0} \frac{\sin(kx)}{x} = k and \lim_{x \to 0} \frac{\tan(kx)}{x} = k.

Equate the limit value to

f(0)

to ensure continuity at

x = 0

a^2 - 5 = 4

By definition, a function is continuous at x = c if and only if \lim_{x \to c} f(x) = f(c).

Solve the resulting quadratic equation for the positive constant

a

a = 3

Adding 5 to both sides gives a^2 = 9; taking the principal square root gives a = 3 since a > 0.

Limits and Continuity of Functions