First slide
Introduction to limits
Question

limxx4sin1x+x21+|x|3 is

Moderate
Solution

limxx4sin1x+x21+|x|3

LetL=limxx31+|x|3xsin1x+1x

        =limxx3|x|311+1|x|xxsin1x+1x

now,

limxx3|x|3=limxx3x3=1[|x|=x for x<0]limx1|x|x=0

and limxxsin1x=limy0sinyy=1  where y=1x

From eq (i) L=111+0(1+0)=1

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