Finding $\lim_{x\to\pi}\frac{\sin5x}{x-\pi}$ without using L'Hostpital's rule

Question

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. $$\lim_{x \to \pi} \frac{{\sin5x}}{x(1-\frac{\pi}{x})}.$$

$$\lim_{x \to \pi} \frac{5}{1-\frac{\pi}{x}}$$

Answer 1

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}

Answer 2

$$\lim_{x \to \pi} \cdot \frac{\sin(5x)}{x-\pi} = \lim_{x \to \pi} \frac{\sin(5x-5\pi + \pi)}{x-\pi} = \lim_{x \to \pi} \frac{\sin(5x-5\pi)\cos(\pi) + \sin(\pi)\cos(5x-5\pi)}{x-\pi} = \lim_{x \to \pi} \frac{\sin(5x-5\pi)\cdot -1 + 0}{x-\pi}$$

$h = x-\pi$

$$-\lim_{h \to 0} \frac{\sin(5h)}{h} = -5$$

Answer 3

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 $x=\pi$

Answer 4

This might be similar to some parts of other answers, but this distills what I think is most important: $$ \begin{align} \lim_{x\to\pi}\frac{\sin(5x)-\sin(5\pi)}{x-\pi} &=\left.\frac{\mathrm{d}}{\mathrm{d}x}\sin(5x)\right|_{\large x=\pi}\\ &=5\cos(5\pi)\\[6pt] &=-5 \end{align} $$

Answer 5

$\sin(5x-5π+5π)= $

$\sin[(5(x-π) -π)+3×2π] =$

$\sin[5(x-π) -π] = -\sin5(x-π).$

$\lim_{x \rightarrow π}\dfrac{\sin5x}{x-π}=$

$\lim_{x \rightarrow π}( -1)5\dfrac{\sin5(x-π)}{5(x-π)}= -5.$