Question

does the x stay or do i change the x into a y?

Answer 1

Given:

$\lim_{x\to1}\frac{sin(\pi x)}{x-1}$

1. You need to evaluate it by substituting this value:

$x=1$

Then:

$=\frac{sin(\pi(1))}{1-1}=\frac{0}{0}$

2. Since it has an indeterminate form of zero over zero, you can use L'Hopital Rule and derivate the numerator and the denominator:

- Set up:

$\lim_{x\to1}\frac{\frac{d}{dx}(sin(\pi x))}{\frac{d}{dx}(x-1)}$

- Apply these Derivative Rules:

$\begin{gathered} \frac{d}{dx}(sin(u(x))=cos(u(x))\cdot u^{\prime}(x) \\ \\ \frac{d}{dx}(x^n)=nx^{n-1} \\ \\ \frac{d}{dx}(k)=0 \end{gathered}$

Where "k" is a constant.

Then:

$\lim_{x\to1}\frac{cos(\pi x)\cdot(\pi x)^{\prime}}{1}$$\lim_{x\to1}\frac{\pi cos(\pi x)}{1}$$\lim_{x\to1}\pi cos(\pi x)$

3. Evaluate:

$=\pi cos(\pi(1))=\pi(-1)=-\pi$

Hence, the answer is:

$=-\pi$