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evaluate using l'Hopital's rule lim x tends to 1[x^3-2x^2+x]/[x^5-1]
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on top we have 3x^2-4x+1 on bottom we have 5x^4 the reason we can do f'/g' is because we have f(1)=0/g(1)=0 so we don't have than anymore we can use direct substittion now we have on top 3-4+1=0 on bottom we have 5*1=5 0/5=0 is the limit
3x^2-4x+1/5x^4 would be your new equation using l'Hoptial's rule now plug in 1 to get 3(1)-4+1/5=0
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