Carly Cannon

2022-07-08

Solving this trig limit without L'hopital: $\underset{x\to 1}{lim}\frac{\mathrm{sin}\left(\pi x\right)}{x-1}$
One idea I had was to multiply both sides of the fractions by
$\pi x$
like this
$\underset{x\to 1}{lim}\frac{\mathrm{sin}\left(\pi x\right)\pi x}{\pi {x}^{2}-\pi x}$
to cancel out the sine but then I realized that I could not do that because x tends to 1, and not to 0. What can I do?

Expert

$\underset{h\to 0}{lim}\frac{\mathrm{sin}\left(\pi \left(1-h\right)\right)}{1-1-h}$
$\begin{array}{}\text{(Multiply and divide by pi)}& \underset{h\to 0}{lim}\frac{\mathrm{sin}\left(\pi h\right)\pi }{-\pi h}\end{array}$
$-\pi$

Sylvia Byrd

Expert

$\begin{array}{rcl}\underset{x\to 1}{lim}\frac{\mathrm{sin}\left(\pi x\right)}{1-x}& =& \underset{h\to 0}{lim}\frac{\mathrm{sin}\left(\pi \left(1-h\right)\right)}{h}\\ & =& \underset{h\to 0}{lim}\frac{\mathrm{sin}\left(-\pi h\right)}{h}\\ \text{(common limit)}& & =& \underset{h\to 0}{lim}\frac{-\mathrm{sin}\left(\pi h\right)\pi }{\pi h}& =& -\pi .\end{array}$