Limit as $x\to 2$ of $\frac{\cos(\frac \pi x)}{x-2} $

Question

I am a kid trying to teach myself Calculus in order to prepare for next year.

I have the expression

$$\lim_{x\to 2}\frac{\cos(\frac \pi x)}{x-2} $$

There is a hint that says to substitute t for $(\frac \pi2 - \frac \pi x)$ and WolframAlpha evaluates this expression as $\frac \pi4$. However, I got the answer of $1$. Can someone clarify the steps to solving this problem.

Answer 1

HINT:

$$\cos\dfrac\pi x=\sin\left(\dfrac\pi2-\dfrac\pi x\right)=\sin\dfrac{\pi(x-2)}{2x}$$

Now set $\dfrac{\pi(x-2)}{2x}=y$ and use $\lim_{h\to a}\dfrac{\sin(h-a)}{h-a}=1$

Answer 2

Here is my solution using only $\lim_{x \to 0} \frac{\sin(x)}{x} = 1$, using the methods the questions asks for (the $t$ substitution) $$\lim_{x \to 2} \frac{\cos(\pi/x)}{x-2}$$ $$ = \lim_{x \to 2} \frac{\sin(\frac{\pi}{2} - \frac{\pi}{x})}{x-2}$$ substitute $t= \frac{\pi}{2} - \frac{\pi}{x}$, $\;\;x = \frac{2 \pi}{\pi-2 t}$ $$ = \lim_{t \to 0} \frac{\sin(t)}{\frac{2 \pi}{\pi-2 t}-2}$$ $$ = \lim_{t \to 0} \frac{\sin(t)}{\frac{2 \pi - 2(\pi-2 t)}{\pi-2 t}}$$ $$ = \lim_{t \to 0} \frac{(\pi-2 t)\sin(t)}{2 \pi - 2(\pi-2 t)}$$ Substitute $r = \pi-2 t$, $\;\;t = \frac{\pi-r}{2}$ $$ = \lim_{r \to \pi} \frac{r\sin(\frac{\pi-r}{2})}{2(\pi - r)}$$ $$ = \frac{1}{4}\lim_{r \to \pi} \frac{r\sin(\frac{\pi-r}{2})}{\frac{\pi-r}{2}}$$ Substitute $k = \frac{\pi-r}{2}$, $\;\;r= \pi-2k$ $$ = \frac{1}{4}\lim_{k \to 0} \frac{(\pi-2k)\sin(k)}{k}$$ $$ = \frac{1}{4}\bigg[\lim_{k \to 0} (\pi-2k)*\lim_{k \to 0}\frac{\sin(k)}{k}\bigg]$$ $$=\frac{\pi}{4}$$ I unfortunately only got this after an answer was accepted, as messing around with a limit this much takes a while and involved some trial and error before I could figure out what to do... there might be a way to condense this proof, but I always jump right into these things by making substitution after substitution until I get to a form that I know.

Answer 3

$$\lim_{x\to2}\frac{\cos(π/x)}{x-2}$$ Which leads us to an Indeterminate form i.e.,(0/0) form , so apply L-Hospital rule ,
$$\lim_{x\to2}[-\sin(π/x)\{-π/x^2\}] =\lim_{x\to2} \fracπ{x^2} =\frac{\pi}{4}$$

Answer 4

Let $x=2+y$ with $y\ne 0$.We have $x-2=y$ and $ \pi/x=\pi/(2+y)=\pi/2-\pi y/(4+2 y) . $ So $\cos \pi/x=\sin \pi y/(4+2 y)=(1-f(y)(\pi y/(4+2 y)$ where $\lim_{y\to 0} f(y)=0 . $ Hence $(\cos \pi/x)/(2-x)= (1-f(y)(\pi/(4+2 y)).$