I want to find the limits $$\lim_{x\to \pi/2} \frac{\cos x}{x-\pi/2} $$
and
$$\lim_{x\to\pi/4} \frac{\cot x – 1}{x-\pi/4} $$
and
$$\lim_{h\to0} \frac{\sin^2(\pi/4+h)-\frac{1}{2}}{h}$$
without L'Hospital's Rule.
I know the fundamental limits $$\lim_{x\to 0} \frac{\sin x}{x} = 1,\quad \lim_{x\to 0} \frac{\cos x – 1}{x} = 0 $$
Progress
Using $\cos x=\sin\bigg(\dfrac\pi2-x\bigg)$ I got $-1$ for the first limit.
Best Answer
In the first substittute $x=\pi/2+h$ so that when $x\to\pi/2,h\to0$ and similiarly for the other one and use some trigonometric identities to convert first one to sine and second to half angle.