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anonymous
 4 years ago
For which values of c will the limit as x → 5 of the piecewisedefined function f(x) = {ln(x+c) for x>5; sqrt(x4) 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!
anonymous
 4 years ago
For which values of c will the limit as x → 5 of the piecewisedefined function f(x) = {ln(x+c) for x>5; sqrt(x4) 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!

This Question is Closed

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0for 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

freckles
 4 years ago
Best ResponseYou've already chosen the best response.2\[\ln(5+c)=\sqrt{54} \text{ we want left limit equal to right limit} \]

freckles
 4 years ago
Best ResponseYou've already chosen the best response.2\[\ln(5+c)=1 \] You need to solve this

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0I see, that makes sense, thanks... would that mean \[c = e  5\]? That seems ... weird, did I do something wrong? :

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Oh 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 :\

freckles
 4 years ago
Best ResponseYou've already chosen the best response.2Well f(5)=lim x> 5 f(x) in order for it to also be continuous at x=5

freckles
 4 years ago
Best ResponseYou've already chosen the best response.2so what did you get for lim x> 5 f(x) ?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Oh, it's 1, right? So... f(5) = 1 for f(x) to be continuous at x=5?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Ahh, how silly of me, thanks :)
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