Limits with radicals question
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Steven G
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Note that sqrt(x^2)= x is not true!. sqrt( (-5)^2 ) = sqrt(25) = 5 NOT -5.
In fact, sqrt(x^2) = | x |
since 4x^2-x = x^2(4-1/x) we have sqrt(4x^2-x) = sqrt( x^2(4-1/x) ) = sqrt(x^2)* sqrt(4 - 1/x) = | x |*sqrt(4-1/x)
Now recall that if x<0, then | x | = -x. Just like | -7 | = -(-7)
Since x is approaching negative infinity, then x will be negative.
Since x<0 (x is approaching negative infinity), sqrt(4x^2-x) = sqrt( x^2(4-1/x) ) = sqrt(x^2)* sqrt(4 - 1/x) = | x |*sqrt(4-1/x) = -x(sqrt(4-1/x)).
Calculus is easy, it is the algebra that is hard!
In fact, sqrt(x^2) = | x |
since 4x^2-x = x^2(4-1/x) we have sqrt(4x^2-x) = sqrt( x^2(4-1/x) ) = sqrt(x^2)* sqrt(4 - 1/x) = | x |*sqrt(4-1/x)
Now recall that if x<0, then | x | = -x. Just like | -7 | = -(-7)
Since x is approaching negative infinity, then x will be negative.
Since x<0 (x is approaching negative infinity), sqrt(4x^2-x) = sqrt( x^2(4-1/x) ) = sqrt(x^2)* sqrt(4 - 1/x) = | x |*sqrt(4-1/x) = -x(sqrt(4-1/x)).
Calculus is easy, it is the algebra that is hard!