Finding $\lim_{x\to\pi}\frac{\sin5x}{x-\pi}$ without using L'Hostpital's rule I have the following function which needs to be found. Obviously this function is fairly straight to find the limit using L'hospital's method (which will give $-5$). However, I need to find the limit without L'hospital's rule.
$$\lim_{x \to \pi} \frac{\sin5x}{x-\pi}.$$
I've attempted something like this by taking the x out of the denominator, but I'll still get an indeterminate form.
$$\lim_{x \to \pi} \frac{{\sin5x}}{x(1-\frac{\pi}{x})}.$$
$$\lim_{x \to \pi} \frac{5}{1-\frac{\pi}{x}}$$

$\sin(5x-5π+5π)= $
$\sin[(5(x-π) -π)+3×2π] =$
$\sin[5(x-π) -π] = -\sin5(x-π).$
$\lim_{x \rightarrow π}\dfrac{\sin5x}{x-π}=$
$\lim_{x \rightarrow π}( -1)5\dfrac{\sin5(x-π)}{5(x-π)}= -5.$
