1. ## Help Limits

Hii !

I don't know Calculat This Limits :

1. $\displaystyle \lim_{x \to 1^+ } \frac{\sqrt{x^2-9} + \sqrt{x} - \sqrt{3}}{\sqrt{x-3} }$

2. $\displaystyle \lim_{x \to \frac{\Pi }{4}} \frac{1 -\sqrt{2}\cos x}{ 1-\sqrt{2}\sin x }$

Can You Help Me

2. Originally Posted by Perelman
Hii !

I don't know Calculat This Limits :

1. $\displaystyle \lim_{x \to 1^+ } \frac{\sqrt{x^2-9} + \sqrt{x} - \sqrt{3}}{\sqrt{x-3} }$
$\displaystyle x=1$ is a point at which the expression is continuous so just substitute $\displaystyle x=1$ and evaluate (it is complex).

2. $\displaystyle \lim_{x \to \frac{\Pi }{4}} \frac{1 -\sqrt{2}\cos x}{ 1-\sqrt{2}\sin x }$
L'Hopital's rule:

$\displaystyle \lim_{x \to \frac{\pi }{4}} \frac{1 -\sqrt{2}\cos (x)}{ 1-\sqrt{2}\sin (x) }=\lim_{x \to \frac{\pi }{4}} \frac{+\sqrt{2}\sin (x)}{ -\sqrt{2}\cos (x) }$

CB

3. or note that $\displaystyle \frac{1-\sqrt{2}\cos x}{1-\sqrt{2}\sin x}=\frac{1-2\cos ^{2}x}{1-2\sin ^{2}x}\cdot \frac{1+\sqrt{2}\sin x}{1+\sqrt{2}\cos x},$