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where is the function f(c)
c=3, f(x)= (x^2+5)/(x-6) continuous or no?... is the question.
at c=3, yes... the function is defined at x=3 so the limit of f is f(3)
what if x=c is not defined?
if f(x) is not defined at x=c, then the limit can still exist but not necessarily at f(c). also, if there is a vertical asymptote at x=c, then the limit does not exist.
the limit can still exist but not necessarily at f(c)?
yes... for example... |dw:1361168628388:dw| here, the limit of f(x) as x approaches c is a.... NOT f(c)
OHHH so then in that case it would be dicontinuous? When can you tell, or what can you do, to know that something is continuous, but algebraically? step by step and explain?
yes... but in your original function, it is continuous at all x values except at x=6. so in your problem, the limit of f as x approaches c for any value OTHER THAN 6, will be f(c)
oh! so then if x=6 was not an asymptote, but a hole, the limit would be equivalent to f(c)??
***IF*** the function was CONTINUOUS at x=6, then the limit would be f(6)
seems like we're going in circles here but we're not... Continuity of a funtion is defined in terms of limits....
if the "hole" you're referring to is a REMOVABLE discontinuity, then yes, the limit would be f(6) or as you said, f(c)...
yes!!!! ok, thank you!