AP Calculus AB and BC

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

The limit of sin 5x over 3x as x approaches 0 is 5 over 3. Rewrite it as 5 over 3 times sin 5x over 5x, so the inner angle matches the denominator exactly. That second factor is the special trigonometric limit and tends to 1.

lim⁡x→0sin⁡5x3x=53\lim_{x \to 0} \frac{\sin 5x}{3x} = \frac{5}{3}

Settled by matching the inner angle to the denominator.

Making the angle match the denominator

The special limit lim⁡u→0sin⁡uu=1\lim_{u \to 0}\frac{\sin u}{u} = 1 only fires when the thing inside the sine is IDENTICAL to the denominator. Here it is 5x5x inside and 3x3x underneath, so manufacture the match.

sin⁡5x3x=53⋅sin⁡5x5x\frac{\sin 5x}{3x} = \frac{5}{3}\cdot\frac{\sin 5x}{5x}

As x→0x \to 0 so does 5x5x, so the second factor is the special limit with u=5xu = 5x and equals 11.

lim⁡x→0sin⁡5x3x=53⋅1=53\lim_{x \to 0}\frac{\sin 5x}{3x} = \frac{5}{3}\cdot 1 = \frac{5}{3}

The general shape

The limit of sin(ax) over bx as x approaches 0 is a over b. Reading that off directly saves rewriting every time, but it is worth being able to produce the factor rather than only recalling it.

Checking with L'Hopital

Substitution gives 00\frac{0}{0}, so the rule applies and confirms the answer in one pass.

lim⁡x→0sin⁡5x3x  =H  lim⁡x→05cos⁡5x3=53\lim_{x \to 0}\frac{\sin 5x}{3x} \;\overset{\text{H}}{=}\; \lim_{x \to 0}\frac{5\cos 5x}{3} = \frac{5}{3}

The chain rule is what produces the 55 in the numerator, which is the same 55 the algebraic method pulls out front. Two routes, one reason.

The mistakes students make

  • Answering 11 by treating any sin⁡(something)something\frac{\sin(\text{something})}{\text{something}} as the special limit. The inner angle and the denominator have to match.
  • Answering 35\frac{3}{5} by inverting the ratio. The coefficient inside the sine goes on top.
  • Cancelling the xx to get sin⁡53\frac{\sin 5}{3}. You cannot cancel a variable out of a sine argument.

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 sin 5x / 3x as x approaches 0?

It is 53\frac{5}{3}. Rewrite as 53⋅sin⁡5x5x\frac{5}{3}\cdot\frac{\sin 5x}{5x}, and the second factor tends to 11.

What is the general rule?

lim⁡x→0sin⁡(ax)bx=ab\lim_{x \to 0}\frac{\sin(ax)}{bx} = \frac{a}{b}, provided both aa and bb are nonzero constants.

Can I use L'Hopital's rule?

Yes, the form is 00\frac{0}{0}. One pass gives 5cos⁡5x3→53\frac{5\cos 5x}{3} \to \frac{5}{3}.