AP Calculus AB and BC

Limit of tan(5x)/x as x Approaches 0

Q: Why does cosine not contribute?

Because \cos 5x \to 1 as x \to 0, so it multiplies the answer by 1. It matters for the shape of the graph away from zero, not for this limit.

The limit of tangent of 5x over x as x approaches zero is five. Tangent over its own argument tends to one just as sine does, so matching the denominator to 5x leaves a factor of five behind.

\lim_{x \to 0} \frac{\tan 5x}{x} = 5

Settled by matching the inner angle, leaving a stray factor.

Tangent behaves like sine near zero

Since $\tan u = \frac{\sin u}{\cos u}$ and $\cos u \to 1$ , the standard limit carries over: $\lim_{u \to 0}\frac{\tan u}{u} = 1$ .

\frac{\tan 5x}{x} = 5 \cdot \frac{\tan 5x}{5x} \longrightarrow 5 \cdot 1 = 5

Multiplying and dividing by 5 is the whole technique: make the denominator match what is inside, and account for it with a constant out front.

Why this works, geometrically

Near zero all three of $\sin u$ , $\tan u$ and $u$ agree to first order. The sine curve sits just below the line $y = u$ and the tangent curve just above it, and all three become indistinguishable as $u$ shrinks.

That is why both standard limits equal 1, and why the coefficient inside is the only thing that survives.

Not sure which technique a limit wants?

The Limit Method Chooser walks the decision from direct substitution through factoring, the conjugate, and L'Hopital, and says why each one applies or fails.

Frequently asked questions

Is the general rule tan(kx)/x tends to k?

Yes, for any constant k, and the same holds with sine in place of tangent. Both reduce to the standard limit after matching the denominator.

Why does cosine not contribute?

Because $\cos 5x \to 1$ as $x \to 0$ , so it multiplies the answer by 1. It matters for the shape of the graph away from zero, not for this limit.