How can I find the limit 0/0 when using direct substitution?

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Discussion Overview

The discussion revolves around finding limits that result in the indeterminate form 0/0 when using direct substitution. Participants explore various methods for resolving these limits, particularly through algebraic manipulation and the use of conjugates.

Discussion Character

  • Homework-related
  • Mathematical reasoning
  • Exploratory

Main Points Raised

  • One participant expresses difficulty with limits that yield 0/0 and describes their understanding of limits involving quadratic equations, suggesting that factoring allows for cancellation.
  • Another participant questions the computation of ##\frac{2}{\sqrt{2}}##, indicating that there may be an error in the calculations presented.
  • A participant describes their method of multiplying by the conjugate to resolve the limit but realizes an error in their calculations after further discussion.
  • There is a suggestion that multiplying by the conjugate is a useful technique for limits of this nature, with participants agreeing on its utility.
  • One participant proposes a substitution method, setting ##2h=u##, to potentially simplify the limit expression and relate it to derivatives.

Areas of Agreement / Disagreement

Participants generally agree that multiplying by the conjugate is a helpful technique for resolving limits that yield 0/0. However, there remains uncertainty regarding specific calculations and methods, with no consensus on the best approach for all cases.

Contextual Notes

Participants express varying levels of understanding and confidence in their methods, indicating potential gaps in foundational knowledge or computational skills. Some steps in the calculations are not fully resolved, and assumptions about the applicability of certain techniques are not universally accepted.

Who May Find This Useful

Students struggling with limits in calculus, particularly those encountering indeterminate forms, may find this discussion beneficial.

Cal124
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I am really struggling with limits at the moment. Any help would be great! Thanks to anyone in advance if they take the time to read the rest of this.
Basically i am struggling with finding the limit when using direct substitution provides 0/0
I (think) am fine with limits that involve quadratic equations, you factorise which should allow some canceling out, and direct substitution should provide an answer.

Just to check
Lim t->(-1) [(1/(t+1)) - (1/(t^2+3t+2)]
factorizing the t^2+3t+2 i can cancel out to 1/(t+2) which with direct substitution the limit = 1/1 =1

but my problem is when i encounter limits such as
lim h->0 (sqrt(2+2h)-sqrt(2)) / h

I did try multiplying it by the conjugate (sqrt(2+2h)-sqrt(2)) / (sqrt(2+2h)+sqrt(2)) but this gave me 2/sqrt(2) and putting the original into worlfram alpha the answer is apparently 1/sqrt(2)

I'm just not sure on the procedure when dealing with this king of limit. any help would be greatly appreciated!
Thanks
 
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Tell us how you obtained ##\frac{2}{\sqrt{2}}##. Your method is good, but there is likely a computation error.
 
micromass said:
Tell us how you obtained ##\frac{2}{\sqrt{2}}##. Your method is good, but there is likely a computation error.
Hopefully this will be okay to follow apologies if not, not sure how to write equations on here yet.
>>
[sqrt(2+2h) - sqrt(2)]/h . [sqrt(2+2h)+sqrt(2)]/sqrt(2+2h)+sqrt(2)
>>
2+2h+[sqrt(2)sqrt(2+2h)]-[sqrt(2)sqrt(2+2h)]-2
h.sqrt(2+2h)+sqrt(2)
>>
cancels to
2h
h.[sqrt(2+2h)+sqrt(2)]
>>
giving, as h->0 & canceling the h's
2
sqrt(2+2.0)+sqrt(2)
Ive realized my error, thanks a lot i will continue just in case
>>
2/sqrt(2)+sqrt(2) = sqrt(2)/2

just for clarification, is it a rule of thumb in this sort of limit to multiply by the conjugate?
 
Nice work!

Cal124 said:
just for clarification, is it a rule of thumb in this sort of limit to multiply by the conjugate?

Yes, it is a trick that works in a lot of cases.
 
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micromass said:
Nice work!
Yes, it is a trick that works in a lot of cases.

Thanks! Much appreciated.
 
Have you tried setting ##2h=u## to make the expression look more like a derivative?
 
WWGD said:
Have you tried setting ##2h=u## to make the expression look more like a derivative?
It looks like it could be a derivative just as it is.

ƒ'(1) , where ƒ(x) = √(2x) .
 

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