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Mathematics 16 Online
OpenStudy (anonymous):

Simple Piecewise Function, when is it continuous?

OpenStudy (anonymous):

hartnn (hartnn):

it'll be continuous when \(\large \lim \limits_{x\rightarrow -1} \dfrac{x^2+5x+4}{x+1}=a\) know how to find that limit ?

OpenStudy (anonymous):

Oh i set them equal to each other? I never really understood limits, and continuity with piecewise functions.

OpenStudy (anonymous):

Would i factor it to (x+1)(x+4)/(x+1). They would cancel out, and i would be left with (x+4)?

hartnn (hartnn):

i took the limit to -1 and then equated them and yes, that would be correct,

OpenStudy (anonymous):

Alright, so where do i go from there?

hartnn (hartnn):

\(\large \lim \limits_{x\rightarrow -1} x+4=a \) just plug in x= -1 !

OpenStudy (anonymous):

Its undefined... hence the function saying x cannot = -1, thats why we're trying to find a.

OpenStudy (anonymous):

ooh into that.

OpenStudy (anonymous):

so a=3

hartnn (hartnn):

yes! :)

OpenStudy (anonymous):

Thanks : )

hartnn (hartnn):

welcome ^_^

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