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

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') =...

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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` .

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