De-Rationalizing a Surd Fraction: Step-by-Step Guide

[lkh1986]

Homework Statement

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.

Homework Equations

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.

 

[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.

 

[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}

 

[arildno]

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

 

[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. :)

 

[arildno]

 

[lkh1986]

Another question:

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

The answer given is 1.

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

 

[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?

 

[arildno]

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