Question

For which values of c will the limit as x → 5 of the piecewise-defined function f(x) = {ln(x+c) for x>5; sqrt(x-4) for x<5 exist?

I have no idea where to start; for the limit to exist at x=5, the function has to approach the same value from both sides, right? But the c is confusing me... please help. Thanks!

Answer 1

for example, if c=-5, then there is no limit for ln(x+c), when x-->5

try to plot this stuff and you will see, how the "c" affects

Answer 2

\[\ln(5+c)=\sqrt{5-4} \text{ we want left limit equal to right limit} \]

Answer 3

\[\ln(5+c)=1 \]
You need to solve this

Answer 4

freckles is right

Answer 5

I see, that makes sense, thanks... would that mean \[c = e - 5\]?

That seems ... weird, did I do something wrong? :|

Answer 6

Seems awesome to me

Answer 7

:D
Thanks a lot!

Answer 8

awesome )

Answer 9

Oh no, now I closed it but have another question pertaining to this... probably too late :(

What would I need to do to find the value of f(5) such that f(x) is continuous at x = 5? It's eluding me :\

Answer 10

Well f(5)=lim x-> 5 f(x) in order for it to also be continuous at x=5

Answer 11

so what did you get for lim x-> 5 f(x) ?

Answer 12

Oh, it's 1, right? So... f(5) = 1 for f(x) to be continuous at x=5?

Answer 13

Yep yep.

Answer 14

Ahh, how silly of me, thanks :)