Calculate $$\lim_{x\to\frac{\pi}{2}}\frac{\cos x}{x-\frac{\pi}{2}}$$ by relating it to a value of $(\cos x)'$.
The answer is available here (pdf) at 1J-2.
However, I can't seem to make sense of what is actually being done here.
Calculate $$\lim_{x\to\frac{\pi}{2}}\frac{\cos x}{x-\frac{\pi}{2}}$$ by relating it to a value of $(\cos x)'$.
The answer is available here (pdf) at 1J-2.
However, I can't seem to make sense of what is actually being done here.
Applying l'Hopital rule, you have
$$ \lim_{x\to\pi/2}\frac{\cos x}{x-\pi/2}=\lim_{x\to\pi/2}(\cos x)'=\left.(\cos x)'\right|_{x=\pi/2} $$ that in turn becomes $$\left.(\cos x)'\right|_{x=\pi/2}=\left.(-\sin x)\right|_{x=\pi/2}=-1$$
Use the definition of derivative at a point,
$$ f'(x_0) = \lim_{x\rightarrow x_0} \frac{f(x)-f(x_0)}{x-x_0} \,.$$ In your case $f(x)=\cos(x)$ and $x_0=\frac{\pi}{2}$.
HINT : I think you wanna use the notation $y = x - \frac{\pi}{2}$.