# What is the limit x-->0 (sqrt(1+x)-sqrt(1-x))/x ?

sciencesolve | Teacher | (Level 3) Educator Emeritus

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You should substitute 0 for x such that:

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

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

The indetermination 0/0 could be solved using l'Hospital's theorem such that:

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

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

Substituting 0 for x yields:

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

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

Hence, evaluating the given limit under the given conditions yields `lim_(x->0) (sqrt(1+x) - sqrt(1-x))/x = 1` .

tonys538 | Student, Undergraduate | (Level 1) Valedictorian

Posted on

The limit limit `lim_(x->0) (sqrt(1+x)-sqrt(1-x))/x` has to be determined.

Substituting x = 0 in `(sqrt(1+x)-sqrt(1-x))/x` gives the indeterminate result `0/0` .

Let us alter the expression by multiplying it with its conjugate.

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

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

Use the relation `(x - a)(x + a) = x^2 - a^2`

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

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

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

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

Now when we substitute x = 0 the result is `(2)/(sqrt 1 + sqrt 1)` = `2/(2*sqrt 1)` = 1

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