# More help needed manipulating terms

Discussion in 'Differentiation and Integration' started by mmesford, Jan 2, 2022.

1. ### mmesford

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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?

mmesford, Jan 2, 2022

2. ### MathLover1

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

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

=

=

=.......simplify

=

MathLover1, Jan 2, 2022
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3. ### nycmathguy

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Nicely-done.

nycmathguy, Jan 2, 2022
4. ### mmesford

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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}?

mmesford, Jan 3, 2022
5. ### nycmathguy

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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.

nycmathguy, Jan 3, 2022
6. ### mmesford

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

mmesford, Jan 3, 2022
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7. ### nycmathguy

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

nycmathguy, Jan 3, 2022
8. ### nycmathguy

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As x--->1

[2(1) - 3]/[(2•1 + 3)(sqrt{1} + 1)]

Simplify that to find your limit.

nycmathguy, Jan 3, 2022
9. ### MathLover1

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observe the graph

if x=1-> then

Last edited: Jan 3, 2022
MathLover1, Jan 3, 2022
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10. ### mmesford

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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?

mmesford, Jan 3, 2022
11. ### MathLover1

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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

MathLover1, Jan 3, 2022
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12. ### mmesford

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Great, thanks!

mmesford, Jan 3, 2022
13. ### nycmathguy

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Is my reply to this member correct?

nycmathguy, Jan 4, 2022
14. ### MathLover1

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yes

MathLover1, Jan 4, 2022
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15. ### nycmathguy

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

nycmathguy, Jan 4, 2022
16. ### Country Boy

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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.

Country Boy, Jan 5, 2022