# Find the limit

1. Oct 8, 2011

### lkh1986

1. The problem statement, all variables and given/known data

Find the lim of $\frac{\sqrt{6-x}-2}{\sqrt{3-x}-1}$when x approaches 2, without using L'hopital rule or the definition.

2. Relevant equations

3. The attempt at a solution
I try to multiply both the numerator and the denominator to get $\frac{(\sqrt{6-x}-2)(\sqrt{3-x}-1)}{2-x}$, but when I substitute x = 2, the term becomes undefined.

Any suggestion? Thanks.

I am thinking of de'rationalize' the rational surd fraction, but nothing works.

Last edited: Oct 8, 2011
2. Oct 8, 2011

### arildno

1. First off:
There should be a "+1" rather than "-1" in the remaining factor after you myliplied both denominator and numerator with the 2conjugate".

2. secondly:
do the same "conjugate trick", with the surd expression that is contained in the numerator.

3. Oct 8, 2011

### lkh1986

Oops, sorry. Now I have

$\frac{\sqrt{6-x}-2}{\sqrt{3-x}-1}$

=$\frac{(\sqrt{6-x}-2)(\sqrt{3-x}+1)}{(\sqrt{3-x}-1))(\sqrt{3-x}+1)}$

=$\frac{(\sqrt{6-x}-2)(\sqrt{3-x}+1)}{2-x}$

4. Oct 8, 2011

### arildno

Correct.
Now do the same trick with the other surd expression.

5. Oct 8, 2011

### lkh1986

Yay, I get it. I multiply by another conjugate and eventually the (2-x) term will cancel out. Then I will get 2/4 which is equal to 1/2. Thanks so much for the help. :)

$\frac{\sqrt{6-x}-2}{\sqrt{3-x}-1}$

=$\frac{(\sqrt{6-x}-2)(\sqrt{3-x}+1)}{(\sqrt{3-x}-1))(\sqrt{3-x}+1)}$

=$\frac{(\sqrt{6-x}-2)(\sqrt{3-x}+1)}{2-x}$

=$\frac{(\sqrt{6-x}-2)(\sqrt{3-x}+1)(\sqrt{6-x}+2)}{(2-x)(\sqrt{6-x}+2)}$

=$\frac{(\sqrt{3-x}+1)(2-x)}{(2-x)(\sqrt{6-x}+2)}$

=$\frac{\sqrt{3-x}+1}{\sqrt{6-x}+2}$

So if I replace x = 2, I will get 2/4 = 1/2. :)

6. Oct 8, 2011

7. Oct 8, 2011

### lkh1986

Another question:

Find the limit of $\frac{sin^{-1}x}{x}$ when x approaches 0.

Seems like I can't express arcsin explicitly. Any clue on how to start? :)

8. Oct 8, 2011

### lkh1986

Can I use the series expansion?

$\frac{sin^{-1}x}{x}$

=$\frac{x+\frac{X^3}{6}+\frac{3x^5}{40}+...}{x}$

=1+$\frac{x^2}{6}+\frac{3x^4}{40}+...$

Then I can get limit = 1 when I substitute x = 0.

Can we get the same answer without using the series expansion thingy?

9. Oct 8, 2011

### arildno

Or, you could make the substitution x=sin(u), which is valid in the neighbourhood of x=0.