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Question

Suppose limx0f(x)=1 and limx0g(x)=5, then evaluate
limx02f(x)g(x)(f(x)+7)23

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Solution

limx0f(x)=1 & limx0g(x)=5
We know algebra of limits is as
limxa(f(x)±g(x))=limxaf(x)±limxag(x)
limx0[f(x)m.g(x)m]=(limxaf(x))m.(limxag(x))n
limxakf(x)=klimxaf(x)
Now, limx02(f(x)g(x))[f(x)+7]23=2(1)(5)[1+7]23=74

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