Limits and Continuity of a Piecewise Function

[genevieveb]

Homework Statement

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

Homework Equations

none

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.

 

[vela]

You need to use parentheses to make it clear what you're doing. It looks like what you are doing is finding
\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}
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]

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]

Compare lim_x→5f(x) amd f(5) .

 

[vela]

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

 

[genevieveb]

I figured it out, thank you!