AP Calculus AB and BC

Limit of sin(pi x)/(x - 1) at 1 Is -pi

The limit of sine of pi x, over x minus 1, as x approaches 1 is negative pi. Setting u equal to x minus 1 turns the numerator into sine of pi u plus pi, which the shift identity rewrites as minus sine of pi u. The standard sine limit then delivers negative pi.

lim⁡x→1sin⁡(πx)x−1=−π\lim_{x \to 1} \frac{\sin\left(\pi x\right)}{x-1} = -\pi

Settled by substituting u = x - 1, then the special trigonometric limit.

Move the approach point to zero

The standard limit lim⁡u→0sin⁡uu=1\lim_{u \to 0}\frac{\sin u}{u} = 1 is only usable when the angle and the denominator both head to 00. Here the angle heads to π\pi, so shift the variable: let u=x−1u = x-1, which means x=u+1x = u+1 and u→0u \to 0.

sin⁡(πx)x−1=sin⁡(π(u+1))u=sin⁡(πu+π)u\frac{\sin\left(\pi x\right)}{x-1} = \frac{\sin\left(\pi(u+1)\right)}{u} = \frac{\sin\left(\pi u + \pi\right)}{u}

The denominator is now the plain uu the standard limit wants. The numerator still carries an unwanted shift of π\pi inside the sine.

The identity supplies the minus sign

sin⁡(θ+π)=−sin⁡θ\sin\left(\theta+\pi\right) = -\sin\theta
lim⁡u→0−sin⁡(πu)u=−πlim⁡u→0sin⁡(πu)πu=−π\lim_{u \to 0}\frac{-\sin\left(\pi u\right)}{u} = -\pi\lim_{u \to 0}\frac{\sin\left(\pi u\right)}{\pi u} = -\pi

This is the part students lose. Every algebraic step in the substitution is sign neutral, so the minus cannot come from the algebra. It comes from the identity, because xx near 11 puts the angle near π\pi, where the sine curve is falling through zero rather than rising through it.

A quick sign check

Take x a little above 1. Then pi x is a little above pi, where sine is negative, while x minus 1 is positive. Negative over positive is negative, which agrees with an answer of negative pi.

The mistakes students make

The first loses the sign, the last loses the coefficient, and the middle one loses both.

  • Dropping the identity, treating sin⁡(πu+π)\sin\left(\pi u+\pi\right) as sin⁡(πu)\sin\left(\pi u\right), and answering π\pi.
  • Pattern matching straight to sin⁡uu\frac{\sin u}{u} without checking that the angle matches the denominator, and answering 11.
  • Applying the identity but forgetting the π\pi inside the angle, and answering −1-1.

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⁡(πx)x−1\frac{\sin\left(\pi x\right)}{x-1} as x approaches 1?

It is −π-\pi.

Where does the minus sign come from?

From the identity sin⁡(θ+π)=−sin⁡θ\sin\left(\theta+\pi\right) = -\sin\theta. After substituting u=x−1u = x-1 the angle is πu+π\pi u+\pi, and a shift by π\pi flips the sign of sine.

Does L'Hopital's rule confirm the answer?

Yes. Differentiating top and bottom gives πcos⁡(πx)1\frac{\pi\cos\left(\pi x\right)}{1}, and at x=1x = 1 that is πcos⁡π=−π\pi\cos\pi = -\pi.