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Find the limit using L'Hopital 's rule lim x-> infinity x^2 sin(1/x) is it possible to do it using L'Hopital rule?
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rewrite it as \[\huge \frac{ \sin (1/x) }{ 1/x^2 }\] so that L'Hopital's rule can be used.
that's what i did too... it is 0/0.. and then i differetiate it, and it turns out to be a mess o_o
derivative of sin 1/x is -1/x^2 cos (1/x)
umm..shouldnt it be 1/x^-2
-1/x^-2 cos x / (-2x^-3)
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then cancel the common factor (-1/x^-2)
then left with cos (1/x) / 2/x it will become 1/ 0 then
yes, and that is infinity
isnt that undefined..?
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