Limits and Derivatives: Numerical Problems Set (3)

Question

Evaluate $\displaystyle\lim_{x \to 0} \dfrac{\sin 5x - \sin 3x}{x}$.

Solution — Step by Step

Why This Works

The standard limit $\lim_{x \to 0} \sin x / x = 1$ generalises to $\lim_{x \to 0} \sin(kx) / x = k$ via simple substitution: $\sin(kx)/x = k \cdot \sin(kx)/(kx) \to k \cdot 1 = k$. Once we split the original limit into two pieces, each piece is a standard limit and we just plug in the coefficients.

This works because the limit operator distributes over sums when each piece has a finite limit. We must verify both limits exist before splitting.

Alternative Method

Use the sum-to-product identity:

$\sin 5x - \sin 3x = 2\cos 4x \sin x$

Then:

$\lim_{x \to 0}\frac{2\cos 4x \sin x}{x} = 2 \cdot \cos 0 \cdot \lim_{x \to 0}\frac{\sin x}{x} = 2 \cdot 1 \cdot 1 = 2$

Same answer, slightly fewer pieces.

Common Mistake

Some students plug in $x = 0$ into $\sin 5x - \sin 3x$, get $0 - 0 = 0$, and conclude the answer is $0/0 = 0$. That’s wrong — $0/0$ is indeterminate, not zero. We must manipulate before evaluating.