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

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ccc | eNotes Newbie

Posted October 27, 2013 at 2:51 AM via web

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

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Zaca | TA , Undergraduate | Salutatorian

Posted October 27, 2013 at 3:18 AM (Answer #1)

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When there is a square root in the numerator, a common way to get rid of it is multiplying by the conjugate:

`lim_(x->0)((sqrt(x+4)-2)/x) * ((sqrt(x+4)+2)/(sqrt(x+4)+2))`

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

The fours cancel and the "x" on the top and bottom cancel.` `

` `

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

` `At this point we can plug in zero, without having the denominator go to zero:

`1/(sqrt(0+4)+2) = 1/4`

` `

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

` `

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aruv | High School Teacher | Valedictorian

Posted October 27, 2013 at 3:56 AM (Answer #2)

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

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

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

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

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

`=1/(sqrt(4)+2)`

`=1/(2+2)`

`=1/4`

``

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