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.
$$\lim_{x \to \pi} \frac{{\sin5x}}{x(1-\frac{\pi}{x})}.$$
$$\lim_{x \to \pi} \frac{5}{1-\frac{\pi}{x}}$$
 A: $$\lim_{x \to \pi}  \cdot \frac{\sin(5x)}{x-\pi} =   \lim_{x \to \pi} \frac{\sin(5x-5\pi + \pi)}{x-\pi} =  \lim_{x \to \pi} \frac{\sin(5x-5\pi)\cos(\pi) + \sin(\pi)\cos(5x-5\pi)}{x-\pi} = \lim_{x \to \pi} \frac{\sin(5x-5\pi)\cdot -1 + 0}{x-\pi}$$
$h = x-\pi$
$$-\lim_{h \to 0} \frac{\sin(5h)}{h} = -5$$
A: This might be similar to some parts of other answers, but this distills what I think is most important:
$$
\begin{align}
\lim_{x\to\pi}\frac{\sin(5x)-\sin(5\pi)}{x-\pi}
&=\left.\frac{\mathrm{d}}{\mathrm{d}x}\sin(5x)\right|_{\large x=\pi}\\
&=5\cos(5\pi)\\[6pt]
&=-5
\end{align}
$$
A: These kind of problems are typically introduced around the time that it is shown that
$$ \lim_{t\to 0} \frac{\sin(t)}{t} = 1. $$
The trick is to find a way to make that limit appear.  In the current context, the first thing that comes to my mind is making it more explicit that the denominator goes to zero as $x \to \pi$.  We can do this by replacing $x-\pi$ with another variable:
\begin{align}
\lim_{x\to \pi} \frac{\sin(5x)}{x-\pi}
&= \lim_{y \to 0} \frac{\sin(5(y+\pi))}{y} && \text{(set $y=x-\pi$)}
\end{align}
At this point, a little bit of algebraic jiggery-pokery seems necessary to make it a little easier to see what is going on.  Remember that the ultimate goal is to get the denominator to match the argument of the sine function.
\begin{align}
\lim_{y \to 0} \frac{\sin(5(y+\pi))}{y}
&= \lim_{y \to 0} \frac{\sin(5y + 5\pi)}{y} \\
&= \lim_{y \to 0} \frac{\sin(5y)\cos(5\pi) + \cos(5y)\sin(5\pi)}{y} && \text{(angle addition formula)}\\
&= \lim_{y \to 0} \frac{-\sin(5y)}{y} && \text{($\cos(5\pi) = -1$, $\sin(5\pi) = 0$)} \\
&= \lim_{y\to 0} \frac{-\sin(5y)}{y} \frac{5}{5} \\
&= -5\lim_{y\to 0} \frac{\sin(5y)}{5y} \\
&= -5. && \left(\text{since $\lim_{t\to 0} \frac{\sin(t)}{t} = 1$}\right)
\end{align}
A: $\sin(5x-5π+5π)= $
$\sin[(5(x-π) -π)+3×2π] =$
$\sin[5(x-π) -π] = -\sin5(x-π).$
$\lim_{x \rightarrow π}\dfrac{\sin5x}{x-π}=$
$\lim_{x \rightarrow π}( -1)5\dfrac{\sin5(x-π)}{5(x-π)}= -5.$
A: Since $\sin(5\pi)=0$:
$$L=\lim_{x \to \pi} \frac{\sin5x}{x-\pi}=\lim_{x \to \pi} \frac{\sin5x-\sin(5\pi)}{x-\pi}=\lim_{x \to \pi} 2{\cos(5(x+\pi)/2)}\frac {\sin(5(x-\pi)/2)}{x-\pi}$$
$$L=-5\lim_{x \to \pi} \frac {\sin(5(x-\pi)/2)}{\frac {5(x-\pi)}{2}}=-5$$
Note that:
$\lim_{x \to \pi} \frac{\sin5x-\sin(5\pi)}{x-\pi}=5\lim_{x \to \pi} \frac{\sin5x-\sin(5\pi)}{5x-5\pi}$  is just the derivative so it is $5\cos(5x)$ with $x=\pi$
