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Limits and Continuity of a Piecewise Function

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genevieveb
#1
Oct15-11, 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(x2-16)-3/(x-5) 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)(x-4))-3/(x-5) * sqrt((x+4)(x-4))+3/sqrt((x+4)(x-4))+3
lim x->5 (x+4)(x-4)-9/(x-5)[sqrt((x+4)(x-4))-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.
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vela
#2
Oct15-11, 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)(x-4)}-3}{x-5}\cdot \frac{\sqrt{(x+4)(x-4)}+3}{\sqrt{(x+4)(x-4)}+3}[/tex]
My first question to you is, why? I don't see why you have the x-5 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?
genevieveb
#3
Oct15-11, 07:27 PM
P: 5
Sorry! I copied the question wrong the first time. I fixed it. It is all over (x-5) in the first part of the function. Sorry about that.

SammyS
#4
Oct15-11, 07:49 PM
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Limits and Continuity of a Piecewise Function

Compare limx→5f(x) amd f(5) .
vela
#5
Oct15-11, 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).
genevieveb
#6
Oct15-11, 09:06 PM
P: 5
I figured it out, thank you!


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