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Calculus1 22 Online
OpenStudy (wesdg1978):

lim x->0 (x^2-2x+sinx/x) How do I find the limit using: lim x=0 sinx/x=1

hartnn (hartnn):

just separate out the limits, lim x->0 (x^2-2x+sinx/x) = lim x->0 (x^2) - lim x->0 (2x) +lim x->0 (sinx/x) the last limit is =1 for others, just plug in x=0

OpenStudy (wesdg1978):

So the answer is simply 1?

hartnn (hartnn):

or was it (x^2-2x+sin x) / x ? if its (x^2-2x+sinx/x) then yes, limit =1

OpenStudy (wesdg1978):

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