SOLUTION: Please tell me if my process is right or wrong.
lim x->-∞ √(6x^6-1)/(5x^3+1)
for this since it's infinity then -1 and +1 can be disregarded. And therefore I am
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Expressions-with-variables
-> SOLUTION: Please tell me if my process is right or wrong.
lim x->-∞ √(6x^6-1)/(5x^3+1)
for this since it's infinity then -1 and +1 can be disregarded. And therefore I am
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Question 1001509: Please tell me if my process is right or wrong.
lim x->-∞ √(6x^6-1)/(5x^3+1)
for this since it's infinity then -1 and +1 can be disregarded. And therefore I am left with:
lim x->-∞ √(6x^6)/(5x^3)
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:
lim x->-∞ (√(6)√(x^6))/(5x^3)
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?
Then the problem becomes:
lim x->-∞ √(6)*|x^3| / (5x^3)
then
lim x->-∞ √(6)*|-∞^3| / (5(-∞)^3)
since its abs value w/e the value is it will be positive. therefore:
lim x->-∞ √(6)(∞)^3/(5(-∞)^3)
so the ∞'s cancel to be -1 and the √(6)/5 remain constant.
lim x->-∞ - √(6)/5
Please explain out the algebra confusion
Thank you Answer by Fombitz(32388) (Show Source):
You can put this solution on YOUR website! The output of the numerator is always positive.
The domain is limited because of the square root. and
So when, , the denominator is positive, so when heads towards , the function value heads to
and when, , the denominator is negative, so when heads towards , the function values heafs to
