+199 I was just reviewing my calc book when I stumbled across this problem.
Any thoughts?
limx⟶∞√x4−10x−√x4−5x2+75
+118703 \lim_{x\longrightarrow\infty}\frac{\sqrt{x^4-10x}-\sqrt{x^4-5x^2+7}}{5}
limx→∞√x4−10x−√x4−5x2+75=limx→∞(√x4−10x−√x4−5x2+7)(√x4−10x+√x4−5x2+7)5(√x4−10x+√x4−5x2+7)=limx→∞x4−10x−(x4−5x2+7)5(√x4−10x+√x4−5x2+7)=limx→∞5x2−10x−75(√x4−10x+√x4−5x2+7)=limx→∞5x2−10x−75(√x4−10x)(1+√x4−5x2+7√x4−10x)
=15×limx→∞5x2−10x−7(√x4−10x)×limx→∞1(1+√x4−5x2+7√x4−10x)
NOW I will look at each of these limits seperately.
limx→∞5x2−10x−7(√x4−10x)=limx→∞√(5x2−10x−7)2x4−10xexpanding gives=limx→∞√25x4−100x3+30x2+140x+49x4−10xDividing top and bottom by x^4 we get=limx→∞√25−100x+30x2+140x3+49x41−10x3=√25=5-------------------------------------------------------
limx→∞1(1+√x4−5x2+7√x4−10x) =1limx→∞(1+√x4−5x2+7√x4−10x)=1limx→∞(√x4−5x2+7√x4−10x)+1=1limx→∞√(x4−5x2+7x4−10x)+1=1limx→∞√(1−5x2+7x41−10x3)+1=1limx→∞√(1−5x2+7x41−10x3)+1=12
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SO what do we have now.
limx→∞√x4−10x−√x4−5x2+75=15×5×12=12
+26396 Limits
I was just reviewing my calc book when I stumbled across this problem.
limx→∞√x4−10x−√x4−5x2+75
limx→∞√x4−10x−√x4−5x2+75=15⋅limx→∞(√x4−10x−√x4−5x2+7)=15⋅limx→∞(√x4−10x−√x4−5x2+7)(√x4−10x+√x4−5x2+7)(√x4−10x+√x4−5x2+7)=15⋅limx→∞(x4−10x−(x4−5x2+7)√x4−10x+√x4−5x2+7)=15⋅limx→∞(5x2−10x−7√x4−10x+√x4−5x2+7)⋅x2x2=15⋅limx→∞(5x2−10x−7x2√x4−10x+√x4−5x2+7x2)=15⋅limx→∞(5x2−10x−7x2√x4−10xx4+√x4−5x2+7x4)=15⋅limx→∞(5−10x−7x2√1−10x3+√1−5x2+7x4)=15⋅(5−0−0√1−0+√1−0+0)=15⋅(51+1)=15⋅52=12
