Solve the limit $\lim_{x\to1}\frac{\cos(\frac{\pi}{2}x)}{x-1}$ without l'hospital's rule.

Question

Hey could anyone help with this? please show steps, also do not use l'hopital's rule

$$\lim_{x\to1}\frac{\cos(\frac{\pi}{2}x)}{x-1}$$

Answer 1

This is the derivative of the function $f(x) = \cos(\pi x/2 )$ at $x = 1$.

From the definition of the derivative, \begin{align} f'(1) &= \lim_{y \to 1} \, (f(y) -f(1))/(y - 1). \end{align} Note that $f(1) = 0$.

Answer 2

littleO's hint should be sufficient, but here's an alternative: Do the substitution $y\leadsto x-1$, so that your limit takes the form $$\lim_{y\to0}\dfrac{\cos\left(\tfrac\pi2(y+1)\right)}{y}.$$ Now use the angle-addition formula and proceed.

Answer 3

Let $\theta = \frac{\pi}{2}(x - 1) \iff \frac{\pi}{2}x = \theta + \frac{\pi}{2}$ so that $\theta \to 0$ as $x \to 1$. Then we have: $$ \lim_{x\to 1} \frac{\cos(\frac{\pi}{2}x)}{x - 1} = \lim_{\theta \to 0} \frac{\cos(\theta + \frac{\pi}{2})}{\frac{2}{\pi}\theta} = \frac{\pi}{2}\lim_{\theta \to 0} \frac{\sin(\frac{\pi}{2} - (\theta + \frac{\pi}{2}))}{\theta} = -\frac{\pi}{2}\lim_{\theta \to 0} \frac{\sin(\theta )}{\theta} = -\frac{\pi}{2} $$

Answer 4

Starting from definition, the Taylor expansion of $\cos(ax)$ built around $y=b$ is given by $$\cos(ax)=\cos (a b)-a (x-b) \sin (a b)-\frac{1}{2} (x-b)^2 \left(a^2 \cos (a b)\right)+O\left((x-b)^3\right)$$ Make $a=\frac{\pi}{2}$ and $b=1$ and obtain, around $x=1$ $$\cos(\frac{\pi}{2}x)=-\frac{1}{2} \pi (x-1)+O\left((x-1)^3\right)$$

I am sure that you can take from here.