AP Calculus AB and BC

Limit of arcsin x / x as x Approaches 0 Is 1

The limit of arcsin x over x as x approaches 0 is 1. Setting the angle equal to arcsin x turns the quotient into the angle over its own sine, which is the reciprocal of the special trigonometric limit and tends to 1.

lim⁡x→0arcsin⁡xx=1\lim_{x \to 0} \frac{\arcsin x}{x} = 1

Settled by substituting the angle, then the special trigonometric limit.

Turning it into a familiar limit

Let θ=arcsin⁡x\theta = \arcsin x, so that x=sin⁡θx = \sin\theta. As x→0x \to 0 the angle θ→0\theta \to 0 too, because arcsine is continuous and sends 00 to 00.

arcsin⁡xx=θsin⁡θ\frac{\arcsin x}{x} = \frac{\theta}{\sin\theta}

That is exactly the reciprocal of the special trigonometric limit, so it tends to 11.

lim⁡x→0arcsin⁡xx=lim⁡θ→0θsin⁡θ=1\lim_{x \to 0}\frac{\arcsin x}{x} = \lim_{\theta \to 0}\frac{\theta}{\sin\theta} = 1

Why the substitution is legal

Changing the variable inside a limit requires the new variable to approach its own target as the old one approaches its. Arcsine is continuous at 0 and sends 0 to 0, so it does.

The L'Hopital route

The form is 00\frac{0}{0}, and the derivative of arcsine finishes it in one line.

lim⁡x→0arcsin⁡xx  =H  lim⁡x→011−x21=11−0=1\lim_{x \to 0}\frac{\arcsin x}{x} \;\overset{\text{H}}{=}\; \lim_{x \to 0}\frac{\frac{1}{\sqrt{1-x^{2}}}}{1} = \frac{1}{\sqrt{1-0}} = 1

The same is true of arctan⁡x\arctan x over xx, whose derivative 11+x2\frac{1}{1+x^{2}} is also 11 at the origin. Near zero all three of sin⁡x\sin x, arcsin⁡x\arcsin x and arctan⁡x\arctan x behave like xx itself.

The mistakes students make

  • Reading arcsin⁡x\arcsin x as 1sin⁡x\frac{1}{\sin x}. The inverse function is not the reciprocal; 1sin⁡x\frac{1}{\sin x} is csc⁡x\csc x.
  • Forgetting the domain. Arcsine only accepts inputs between −1-1 and 11, so this limit is only meaningful near 00.
  • Answering 00 because the numerator goes to 00. So does the denominator, at the same rate.

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 arcsin x / x as x approaches 0?

It is 11. Substituting θ=arcsin⁡x\theta = \arcsin x turns it into θsin⁡θ\frac{\theta}{\sin\theta}.

Is arcsin x the same as 1 / sin x?

No. arcsin⁡x\arcsin x is the inverse function; 1sin⁡x\frac{1}{\sin x} is csc⁡x\csc x. They are completely different.

What about arctan x / x?

Also 11, for the same reason: near the origin arctan⁡x\arctan x behaves like xx, and its derivative at 00 is 11.