AP Calculus AB and BC

Limit of sin 3x / sin 5x as x Approaches 0 Is 3/5

Q: Is there a shortcut for sin(ax)/sin(bx)?

Yes: the limit as x \to 0 is \frac{a}{b} for nonzero b. Divide top and bottom by x and each sine matches its own special limit, leaving \frac{a \cdot 1}{b \cdot 1}.

Q: Does L'Hopital's rule give the same answer?

It does. The form is \frac{0}{0}, and differentiating gives \lim_{x \to 0} \frac{3\cos 3x}{5\cos 5x} = \frac{3}{5}, since both cosines tend to 1. The chain rule factors 3 and 5 carry the result.

Q: What is the limit of tan 3x / sin 5x?

Still \frac{3}{5}. Writing \tan 3x = \frac{\sin 3x}{\cos 3x} adds a factor of \frac{1}{\cos 3x}, which is continuous at 0 and contributes 1.

The limit of sin 3x over sin 5x as x approaches 0 is 3/5, or 0.6. Substitution gives 0/0. Divide numerator and denominator by x, then match each piece to the special sine limit: the top tends to 3 and the bottom to 5. The answer is the ratio of the two coefficients.

\lim_{x \to 0} \frac{\sin 3x}{\sin 5x} = \frac{3}{5}

Settled by the special trigonometric limit applied twice.

Split it into two special limits

Neither sine is over an $x$ , so manufacture one. Dividing numerator and denominator by $x$ leaves the value unchanged for every $x \neq 0$ , which is all a limit inspects.

\frac{\sin 3x}{\sin 5x} = \frac{\frac{\sin 3x}{x}}{\frac{\sin 5x}{x}}

Fix each argument the usual way, the top with a factor of $\frac{3}{3}$ and the bottom with $\frac{5}{5}$ .

= \frac{3 \cdot \frac{\sin 3x}{3x}}{5 \cdot \frac{\sin 5x}{5x}}

Both inner fractions tend to 1, since $3x \to 0$ and $5x \to 0$ as $x \to 0$ . The quotient law applies because the denominator's limit is 5, which is not zero.

\lim_{x \to 0} \frac{\sin 3x}{\sin 5x} = \frac{3 \cdot 1}{5 \cdot 1} = \frac{3}{5}

Why direct substitution fails

Substituting gives $\frac{\sin 0}{\sin 0} = \frac{0}{0}$ . Two vanishing sines look symmetric, which tempts an answer of 1, but the form reports only that both collapse, not how fast.

The rates are what differ. Near zero $\sin 3x \approx 3x$ and $\sin 5x \approx 5x$ , so the quotient behaves like $\frac{3x}{5x} = \frac{3}{5}$ . The $x$ cancels out of that estimate, which is why the limit is a fixed ratio rather than 0 or infinity.

$x$	$\frac{\sin 3x}{\sin 5x}$
0.1	0.616405
0.01	0.600160
0.001	0.600002

Common mistakes

Answering 1 because both parts are sines. Matching function names is not matching rates.
Reading the coefficients off backwards as $\frac{5}{3}$ . The numerator's coefficient stays on top.
Treating $\sin 3x \approx 3x$ as an identity. It is a small-angle statement, valid only in the limit at 0.
Exporting the shortcut away from zero. The ratio of coefficients answers this question only as $x \to 0$ ; at $x = \frac{\pi}{5}$ the denominator is 0 while the numerator is not, and the expression has a vertical asymptote there.
Dropping chain rule factors under L'Hopital. The rule gives $\frac{3\cos 3x}{5\cos 5x} \to \frac{3}{5}$ , and the 3 and 5 on the outside are the whole answer.

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 there a shortcut for sin(ax)/sin(bx)?

Yes: the limit as $x \to 0$ is $\frac{a}{b}$ for nonzero $b$ . Divide top and bottom by $x$ and each sine matches its own special limit, leaving $\frac{a \cdot 1}{b \cdot 1}$ .

Does L'Hopital's rule give the same answer?

It does. The form is $\frac{0}{0}$ , and differentiating gives $\lim_{x \to 0} \frac{3\cos 3x}{5\cos 5x} = \frac{3}{5}$ , since both cosines tend to 1. The chain rule factors 3 and 5 carry the result.

What is the limit of tan 3x / sin 5x?

Still $\frac{3}{5}$ . Writing $\tan 3x = \frac{\sin 3x}{\cos 3x}$ adds a factor of $\frac{1}{\cos 3x}$ , which is continuous at 0 and contributes 1.