Seeking help for limit math problem

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Homework Help Overview

The discussion revolves around evaluating the limit of the expression (sqrt(x^2+5) - 3)/(x-2) as x approaches 2. Participants are exploring the implications of substituting values directly into the expression and the resulting zero denominator.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the substitution method and its failure due to a zero denominator. There are questions about the validity of the original substitution and the methods learned to handle such cases. Some participants suggest using the conjugate to simplify the expression.

Discussion Status

The discussion is active, with participants sharing their attempts and questioning each other's reasoning. Some guidance has been offered regarding the use of the conjugate, but there is no explicit consensus on the correct approach yet.

Contextual Notes

Participants mention instructions from their course regarding the multiplication of the numerator and denominator by the conjugate, indicating a structured approach to the problem. There is also acknowledgment of potential arithmetic errors in the participants' calculations.

teffy3001
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lim (sqrt(x^2+5) - 3)/(x-2)
x->2


i hope that kinda makes sense...

anyways, my question is that I've been doing this and i keep getting 0 in the denominator in the answer...after subsituting everything i got 4/ sqrt(9) - 3 = 4/0
so then is this equation just not possible?
 
Last edited:
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teffy3001 said:
after subsituting everything i got 4/ sqrt(9) - 3 = 4/0

How did you get this? That doesn't appear to bear any relation to the initial expression you gave.
 
teffy3001 said:
lim (sqrt(x^2+5) - 3)/(x-2)
x->2


i hope that kinda makes sense...

anyways, my question is that I've been doing this and i keep getting 0 in the denominator in the answer...after subsituting everything i got 4/ sqrt(9) - 3 = 4/0
so then is this equation just not possible?

For [tex]\lim_{x \rightarrow 2}\frac{\sqrt{x^2+5}-3}{x-2}[/tex]

Clearly the substitution method will give you a zero denominator!

So, what methods have you learned to try and "avoid" a zero denominator?

Casey

matt grime said:
How did you get this? That doesn't appear to bear any relation to the initial expression you gave.

I think they meant (sqrt(4+5)-3)/0...
Arithmetic errors are always a good start:wink:
 
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um the course I am taking told me first multiply the numerator and denominator by
sqrt(x^2+5) - 3

i might have done it wrong but then i got x^2+5-9/x-2(sqrt(x^2+5) - 3)
i factored that to x^2-4/x-2(sqrt(x^2+5) - 3)
and then eventually got (x-2)(x+2)/x-2(sqrt(x^2+5) - 3)
canceled out x-2
x+2/sqrt(x^2 + 5) - 3
then subtituted the x for 2
thats how i got that answer...i think that's how the teacher told me to do it...probably made mistakes along the way though..im not sure
 
um the course I am taking told me first multiply the numerator and denominator by
sqrt(x^2+5) - 3

i might have done it wrong but then i got x^2+5-9/x-2(sqrt(x^2+5) - 3)
Where is the interaction term in your numerator? How do you expand (a + b)^2?
 
teffy3001 said:
um the course I am taking told me first multiply the numerator and denominator by
sqrt(x^2+5) - 3

i might have done it wrong but then i got x^2+5-9/x-2(sqrt(x^2+5) - 3)
i factored that to x^2-4/x-2(sqrt(x^2+5) - 3)
and then eventually got (x-2)(x+2)/x-2(sqrt(x^2+5) - 3)
canceled out x-2
x+2/sqrt(x^2 + 5) - 3
then subtituted the x for 2
thats how i got that answer...i think that's how the teacher told me to do it...probably made mistakes along the way though..im not sure

Your on the right track by using the conjugate; however, you need to watch your signs
the conjugate of [tex]\sqrt{x^2+5}-3[/tex] is [tex]\sqrt{x^2+5}+3[/tex]

Casey

PS~Do you understand why the sign needs to be a plus (+) sign?
Hint: [tex](a+b)^2[/tex] does not = [tex]a^2+b^2[/tex]
Casey
 
Last edited:

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