## FrozenCthulhu 3 years ago 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!

1. Gepard

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

2. freckles

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

3. freckles

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

4. Gepard

freckles is right

5. FrozenCthulhu

I see, that makes sense, thanks... would that mean $c = e - 5$? That seems ... weird, did I do something wrong? :|

6. freckles

Seems awesome to me

7. FrozenCthulhu

:D Thanks a lot!

8. Gepard

awesome )

9. FrozenCthulhu

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 :\

10. freckles

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

11. freckles

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

12. FrozenCthulhu

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

13. freckles

Yep yep.

14. FrozenCthulhu

Ahh, how silly of me, thanks :)