# Limits with trigonometric functions without using L'Hospital Rule.

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.

• Hint: $\cos x=\sin\bigg(\dfrac\pi2-x\bigg)$. – Lucian Nov 7 '14 at 3:26
• that helped, i got -1. – Lying Cat Nov 7 '14 at 3:40
• I added another question. – Lying Cat Nov 7 '14 at 5:17

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.

$$\lim_{x\to\dfrac\pi4}\frac{\cot x-1}{x-\dfrac\pi4}$$

$$=\lim_{x\to\dfrac\pi4}\frac{\cot x-\cot\dfrac\pi4}{x-\dfrac\pi4}\ \ \ \ (1)$$

$$(1)=-\lim_{x\to\dfrac\pi4}\frac{\sin\left(x-\dfrac\pi4\right)}{x-\dfrac\pi4}\cdot\frac1{\lim_{x\to\dfrac\pi4}\sin x\sin\dfrac\pi4}=?$$

Alternatively, $(1)=$ $$\frac{d(\cot x)}{dx}_{\text{ at }x=\dfrac\pi4}$$

Similarly, $$\lim_{x\to\dfrac\pi2}\frac{\cos x}{x-\dfrac\pi2}=\lim_{x\to\dfrac\pi2}\frac{\cos x-\cos\dfrac\pi2}{x-\dfrac\pi2}$$

We can use Prosthaphaeresis Formula now