Carly Cannon

Answered

2022-07-08

Solving this trig limit without L'hopital: $\underset{x\to 1}{lim}\frac{\mathrm{sin}(\pi x)}{x-1}$

One idea I had was to multiply both sides of the fractions by

$\pi x$

like this

$\underset{x\to 1}{lim}\frac{\mathrm{sin}(\pi x)\pi x}{\pi {x}^{2}-\pi x}$

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?

One idea I had was to multiply both sides of the fractions by

$\pi x$

like this

$\underset{x\to 1}{lim}\frac{\mathrm{sin}(\pi x)\pi x}{\pi {x}^{2}-\pi x}$

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?

Answer & Explanation

Madden Luna

Expert

2022-07-09Added 9 answers

$\underset{h\to 0}{lim}\frac{\mathrm{sin}(\pi (1-h))}{1-1-h}$

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

$-\pi $

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

$-\pi $

Sylvia Byrd

Expert

2022-07-10Added 6 answers

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

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