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

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The limit `lim_(x->0) (sqrt(x+4) - 2)/x` has to be determined.

If we substitute x = 0 in `(sqrt(x+4) - 2)/x` the result is `0/0` . This is an indeterminate form and we can use l'Hospital's rule to determine the limit by replacing the numerator and denominator by their derivatives.

`(sqrt(4+x) - 2)' = (1/2)*(1/sqrt(x+4))`

`x' = 1`

This gives the limit:

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

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

Substituting x = 0 gives the result `1/(2*sqrt4) = 1/4`

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

`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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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`

` `