More help needed manipulating terms

My basic algebra skills are still letting me down. Trying to find the limit of this term:

[(2x-3)(sqrt{x}-1)]/[2x^2+x-3]

It seems obvious that I need to get the ‘x’ out of the sqrt but I can’t figure out how. I tried multiplying top and bottom by (sqrt{x}+1) but that just over the problem to the denominator. I tried squaring top and bottom but I’m left with (x-2*sqrt{x}+1) in the numerator. I can’t envision anything that frees up the sqrt{x} term. Any ideas?

 

..........factor 2x^2+x-3= (x - 1) (2 x + 3)

=........rationalize , multiply both numerator and denominator by (sqrt(x)+1)

=

=

=.......simplify

=

 

Nicely-done.

 

Thank you, but I don’t understand how this helps. I was able to get this far but don’t I need to get rid of the sqrt{1} in order to calculate the limit?
If I plug 1 into the final form I end up with:
[(2-3)(sqrt{1}-1)]/2+1-3 which simplifies to:
-1/(5*sqrt{1}+5)
If I ignore the sqrt{1} it simplifies to -1/10 which is the answer in the book. But how do I get away with ignoring the sqrt{1}?

 

I have never taken calculus but I understand what she did. What is x tending to? If, for example, x approaches 0, no need to remove the square root from the denominator. Just replace x with whatever x is approaching and simplify.

 

Sorry, I forgot to mention, we’re looking for the limit as x approaches 1. But you have the concept down.

 

Beautiful. Replace every x you see with 1 and simplify.

 

As x--->1

[2(1) - 3]/[(2•1 + 3)(sqrt{1} + 1)]

Simplify that to find your limit.

 

observe the graph





if x=1-> then

 

So it’s legal to just ignore the imaginary part? And the only reason we went to all that trouble was so there wasn’t a zero in the denominator?

 

yes, and the only reason you can face trouble was that x=1 would make the denominator of the original expression equal to zero:

[(2x-3)(sqrt{x}-1)]/[2x^2+x-3]=>if x=1 than [2*1^2+1-3]=2+1-3=0

 

Great, thanks!

 

Is my reply to this member correct?

 

yes

 

Cool. Just posted two more Half-Angle Formulas: 41 and 44.

 

Not that in lim_(x->a) f(x)/g(x),
if neither f(a) nor g(a) is 0, the limit is f(a)/g(a).
if f(a)= 0 but g(a) is not 0, the limit is 0.
if f(a) is not 0 but g(a)= 0, the limit does not exist.

The only 'problem' is if f(a)= 0 and g(a)= 0. If f and g are polynomials (or can be made into polynomials as here) then they must both have factors of (x- a) which can be canceled.