Solving Confusing Limits Homework

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    Confusing Limits
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The discussion focuses on solving two limit problems: limh->0 (sqrt(a+h)-sqrt(a))/h and limx->1 (1-sqrt(x))/(1-x). Participants suggest using the conjugate method to simplify the first limit, which leads to a solution of 1/2. For the second limit, factoring the numerator as (1 - sqrt(x))(1 + sqrt(x)) is recommended, making the solution clearer. The contributors confirm that these methods effectively solve the limits.
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Homework Statement



1. limh->0 (sqrt(a+h)-sqrt(a))/h
2. limx->1 (1-sqrt(x))/(1-x)

Homework Equations



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The Attempt at a Solution



I tried conjugate, etc., but couldn't solve it without L'H's Rule.
 
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Using the conjugate will work. What did you get when you tried that?
 
Thanks I got 1/(2)for one and 1/2 for number 2
 
Last edited:
For the second one, try factoring the numerator like so:

1 - x =\left(1 - \sqrt{x}\right)\left(1 + \sqrt{x}\right)

The solution should then become obvious.

EDIT: And for the first one, multiplying by the conjugate of the numerator is what you need to do.
 
Thanks guys, got them.

Ignore my comment above this.
 
I tried to combine those 2 formulas but it didn't work. I tried using another case where there are 2 red balls and 2 blue balls only so when combining the formula I got ##\frac{(4-1)!}{2!2!}=\frac{3}{2}## which does not make sense. Is there any formula to calculate cyclic permutation of identical objects or I have to do it by listing all the possibilities? Thanks
Since ##px^9+q## is the factor, then ##x^9=\frac{-q}{p}## will be one of the roots. Let ##f(x)=27x^{18}+bx^9+70##, then: $$27\left(\frac{-q}{p}\right)^2+b\left(\frac{-q}{p}\right)+70=0$$ $$b=27 \frac{q}{p}+70 \frac{p}{q}$$ $$b=\frac{27q^2+70p^2}{pq}$$ From this expression, it looks like there is no greatest value of ##b## because increasing the value of ##p## and ##q## will also increase the value of ##b##. How to find the greatest value of ##b##? Thanks
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