What is the numerical value of the limit: \lim_{x \to 0} \frac{1 - \cos(4x)}{x \sin(2x)}? — JAMB UTME

What is the numerical value of the limit:

\lim_{x \to 0} \frac{1 - \cos(4x)}{x \sin(2x)}

The numerical value of the limit is 4.

Applying the double-angle trigonometric identity

1 - \cos(4x) = 2\sin^2(2x)

reduces the expression to

\frac{2\sin(2x)}{x}

. Rewriting this as

4 \cdot \frac{\sin(2x)}{2x}

and taking the limit as

x \to 0

using the standard limit

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

yields 4.

Identify the form of the limit

Direct substitution of

x = 0

gives

\frac{1 - \cos(0)}{0 \cdot \sin(0)} = \frac{0}{0}

, which is an indeterminate form.

Indeterminate forms require algebraic simplification or trigonometric identities before evaluating the limit.

Apply trigonometric identity

Use

1 - \cos(4x) = 2\sin^2(2x)

to rewrite the numerator.

This transforms the numerator into a form containing sine terms matching the denominator.

Simplify the algebraic expression

\lim_{x \to 0} \frac{2\sin^2(2x)}{x\sin(2x)} = \lim_{x \to 0} \frac{2\sin(2x)}{x}

Cancel out the common

\sin(2x)

factor from numerator and denominator for

x \neq 0

Evaluate using the standard trigonometric limit

\lim_{x \to 0} 4 \cdot \frac{\sin(2x)}{2x} = 4 \cdot 1 = 4

Since

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

, setting

\theta = 2x

gives

\lim_{x \to 0} \frac{\sin(2x)}{2x} = 1

Limits of Trigonometric Functions and Indeterminate Forms

Soru