document.write( "Question 1001509: Please tell me if my process is right or wrong.\r
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document.write( "lim x->-∞ √(6x^6-1)/(5x^3+1)\r
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document.write( "for this since it's infinity then -1 and +1 can be disregarded. And therefore I am left with:\r
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document.write( "lim x->-∞ √(6x^6)/(5x^3)\r
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document.write( "The part I don't remember is an algebra rule, when we sqrt the 6x^6 do we break them up or take them together? I think we take them as separate terms 6 and x^6 if so then it's:\r
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document.write( "lim x->-∞ (√(6)√(x^6))/(5x^3)\r
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document.write( "then nothing happens to the √(6), however, another rule I forget is when you squareroot something this is the same as (x^6)^(1/2) right? so then it becomes x^3 however, why does this become an absolute number like so |x^3| I don't remember this rule at all. When you squareroot something whatever it is that becomes absolute?\r
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document.write( "Then the problem becomes:\r
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document.write( "lim x->-∞ √(6)*|x^3| / (5x^3)
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document.write( "then
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document.write( "lim x->-∞ √(6)*|-∞^3| / (5(-∞)^3)\r
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document.write( "since its abs value w/e the value is it will be positive. therefore:
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document.write( "lim x->-∞ √(6)(∞)^3/(5(-∞)^3)\r
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document.write( "so the ∞'s cancel to be -1 and the √(6)/5 remain constant.\r
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document.write( "lim x->-∞ - √(6)/5\r
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document.write( "Please explain out the algebra confusion\r
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document.write( "Thank you \n" );
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Algebra.Com's Answer #618680 by Fombitz(32388)![]() ![]() You can put this solution on YOUR website! The output of the numerator is always positive. \n" ); document.write( "The domain is limited because of the square root. \n" ); document.write( " \n" ); document.write( " \n" ); document.write( " \n" ); document.write( "So when, \n" ); document.write( " \n" ); document.write( "and when, \n" ); document.write( " |