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Limits and Continuity of a Piecewise Function 
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#1
Oct1511, 06:27 PM

P: 5

1. The problem statement, all variables and given/known data
Find a value for k to make f(x) continuous at 5 f(x)= sqrt(x^{2}16)3/(x5) if x cannot equal 5 3x+k when x=5 2. Relevant equations none 3. The attempt at a solution lim x>5 sqrt((x+4)(x4))3/(x5) * sqrt((x+4)(x4))+3/sqrt((x+4)(x4))+3 lim x>5 (x+4)(x4)9/(x5)[sqrt((x+4)(x4))3] And that's as far as I got. I'm not sure what my next step should be or if what I did is wrong. I graphed the function and used a program to solve for the limit which is apparently 5/3, but I couldn't come up with that answer. I would really appreciate any help/suggestions. Thanks. 


#2
Oct1511, 07:14 PM

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You need to use parentheses to make it clear what you're doing. It looks like what you are doing is finding
[tex]\lim_{x \to 5} \frac{\sqrt{(x+4)(x4)}3}{x5}\cdot \frac{\sqrt{(x+4)(x4)}+3}{\sqrt{(x+4)(x4)}+3}[/tex] My first question to you is, why? I don't see why you have the x5 in there. Look up the definition of continuity. You should see you're making this problem harder than intended. Or did you not describe f(x) correctly in the beginning of your post? 


#3
Oct1511, 07:27 PM

P: 5

Sorry! I copied the question wrong the first time. I fixed it. It is all over (x5) in the first part of the function. Sorry about that.



#4
Oct1511, 07:49 PM

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Limits and Continuity of a Piecewise Function
Compare lim_{x→5}f(x) amd f(5) .



#5
Oct1511, 07:56 PM

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Oh, okay. Then you're headed in the right direction. Like SammyS says, calculate the limit and compare it to f(5).



#6
Oct1511, 09:06 PM

P: 5

I figured it out, thank you!



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