AP Calculus AB and BC

Limit of arctan 3x / 5x at 0 Is 3/5

Q: What is the general rule for arctan(ax)/(bx)?

At 0 the limit is \frac{a}{b} for any constants a and b with b \neq 0, since \arctan(ax) behaves like ax for small x.

Q: Is the answer still 3/5 as x goes to infinity?

No. There \arctan 3x \to \frac{\pi}{2} while 5x grows without bound, so the limit is 0.

The limit of arctan 3x over 5x as x approaches 0 is three fifths. Near 0 the arctangent of a small angle is close to the angle itself, so the quotient behaves like 3x over 5x. Rewriting it as three fifths times arctan 3x over 3x makes that exact.

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

Settled by matching the inner angle to the denominator.

Give the arctangent its own angle

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

The quotient $\frac{\arctan 3x}{3x}$ tends to $1$ , because the angle inside the arctangent now matches the denominator exactly. What is left in front is the coefficient ratio $\frac{3}{5}$ .

The family this belongs to

Arcsine and arctangent behave near 0 the way sine and tangent do: the function of a small angle is close to the angle. So arctan(ax) over bx tends to a over b, with b not zero, and the same reading works for arcsin(ax) over bx.

L'Hopital gives the same ratio

Substitution produces $\frac{0}{0}$ , so differentiating the top and the bottom separately is allowed.

\lim_{x \to 0}\frac{\dfrac{3}{1+9x^{2}}}{5} = \frac{3}{5}

The chain rule puts the $3$ on top and the constant multiple rule puts the $5$ underneath. Both routes are really recording the same fact, that $\arctan 3x$ and $3x$ agree to first order at $0$ .

At infinity nothing above applies

As $x \to \infty$ the angle is no longer small. Here $\arctan 3x$ levels off at $\frac{\pi}{2}$ while $5x$ grows without bound, so the quotient collapses to $0$ .

0 \le \left|\frac{\arctan 3x}{5x}\right| \le \frac{\pi}{10x} \longrightarrow 0 \qquad (x > 0)

Read the approach point before choosing the technique. The same expression has two different stories at the two ends.

The mistakes students make

All three come from quoting a memorised limit before checking that the problem matches it.

Quoting the standard limit straight off as $1$ , without checking that the denominator is the inner angle. The denominator here is $5x$ , not $3x$ .
Inverting to $\frac{5}{3}$ . The coefficient from the numerator belongs on top.
Reporting $\frac{\pi}{2}$ , which is the value of $\arctan 3x$ far out along the axis, in a limit taken at $0$ .

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

What is the limit of arctan 3x / 5x as x approaches 0?

It is $\frac{3}{5}$ .

What is the general rule for arctan(ax)/(bx)?

At $0$ the limit is $\frac{a}{b}$ for any constants $a$ and $b$ with $b \neq 0$ , since $\arctan(ax)$ behaves like $ax$ for small $x$ .

Is the answer still 3/5 as x goes to infinity?

No. There $\arctan 3x \to \frac{\pi}{2}$ while $5x$ grows without bound, so the limit is $0$ .