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Mathematics 19 Online
OpenStudy (anonymous):

lim x > 0 (1-sec(x)) / x^2

OpenStudy (anonymous):

do you know l'hospitals

OpenStudy (anonymous):

calc1

OpenStudy (anonymous):

okay, yes or no for l'hospitals

OpenStudy (anonymous):

no

OpenStudy (anonymous):

well, sec is 1/cos

OpenStudy (anonymous):

yea i know

OpenStudy (anonymous):

i can't get pass that

OpenStudy (anonymous):

can you try multipying the top and botttom of the limit by 1+sec(x)

OpenStudy (anonymous):

i did that too

OpenStudy (anonymous):

multiply by the conjugate

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