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` .