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Calculate limit (2x-square root(4x^2+x)) with x--->infinite?

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luvgoj | Student, Undergraduate | (Level 2) Honors

Posted July 21, 2013 at 2:08 PM via web

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Calculate limit (2x-square root(4x^2+x)) with x--->infinite?

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sciencesolve | Teacher | (Level 3) Educator Emeritus

Posted July 21, 2013 at 2:41 PM (Answer #1)

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You need to evaluate the limit, hence, you need to replace `oo` for `x` in function `y = (2x - sqrt(4x^2+x))` , such that:

`lim_(x->oo) (2x - sqrt(4x^2+x)) = (oo - oo)`

You should perform the multiplication and division by the conjugate of the function, such that:

`lim_(x->oo)((2x - sqrt(4x^2+x))(2x + sqrt(4x^2+x)))/(2x + sqrt(4x^2+x))`

Converting the product into a difference of squares, yields:

`lim_(x->oo) (4x^2 - (4x^2 + x))/(2x + sqrt(4x^2+x))`

`lim_(x->oo) (4x^2 - 4x^2 - x)/(2x + sqrt(4x^2+x))`

Reducing duplicate members to numerator, yields:

`lim_(x->oo) (- x)/(2x + sqrt(4x^2+x))`

You need to force factor `x^2` under the square root, such that:

`lim_(x->oo) (- x)/(2x + sqrt(x^2(4+x/x^2)))`

Using the identity `sqrt(x^2) = |x|` yields:

`lim_(x->oo) (- x)/(2x + |x|sqrt(4+1/x))`

Since `x > 0` yields that `|x| = x` , such that:

`lim_(x->oo) (- x)/(2x + x*sqrt(4+1/x))`

You need to force factor x to denominator, such that:

`lim_(x->oo) (- x)/(x(2 + sqrt(4+1/x)))`

Reducing duplicate factors yields:

`lim_(x->oo) (- 1)/(2 + sqrt(4+1/x)) = -1/(2 + sqrt(4 + lim_(x->oo) 1/x))`

`lim_(x->oo) (- 1)/(2 + sqrt(4+1/x)) = -1/(2 + sqrt(4 + 0))`

`lim_(x->oo) (- 1)/(2 + sqrt(4+1/x)) = -1/(2 + 2)`

`lim_(x->oo) (- 1)/(2 + sqrt(4+1/x)) = -1/4`

Hence, evaluating the given limit performing the multiplication by conjugate, yields `lim_(x->oo) (2x - sqrt(4x^2+x)) = -1/4.`

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