What is the limit x-->0 (sqrt(1+x)-sqrt(1-x))/x ?
2 Answers
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
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`