# What is lim x--> 0 [ (sqrt (1+x) -1)/x]

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Here we cannot replace x by 0 to get the result as that yields 0/0 which is indeterminate. Therefore we use L’ Hopital’s Rule which states that if an expression of the form lim x-->0 [f(x)/ g(x)] gives the indeterminate form 0/0, we can use [f’(x)/g’(x)] for x=0 instead of lim x-->0 [f(x)/ g(x)].

Here we have f(x) = (sqrt (1+x) -1) => f’(x) = 1/2 sqrt (1+x)

and g(x) = x => g’(x) = 1

Therefore lim x--> 0 [(sqrt (1+x) -1)/x] = [f’(x)/g’(x)] for x=0

=> [1/2 sqrt (1+x)] / 1 for x =0

=> [1/ 2 sqrt (1+0)] / 1

=> 1/ 2 sqrt 1

=> 1/2

**Therefore the required limit is equal to 1/2.**

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

One way of finding the limit is to use l'Hospital's rule. Another way is to use factorization.

The factorized form of x^2 - y^2 = (x - y)(x + y)

`(sqrt (1+x) -1)/x`

= `(sqrt (1+x) -1)/(1 + x - 1)`

= `(sqrt (1+x) -1)/((sqrt(1 + x))^2 - 1^2)`

= `(sqrt (1+x) -1)/((sqrt(1 + x) - 1)(sqrt(1 + x) + 1))`

= `1/(sqrt(1 + x) + 1)`

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

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

At x = 0, `1/(sqrt (1+x) +1) = 1/(sqrt 1 + 1) = 1/(1 + 1) = 1/2`

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

To find the lt x-->0 {sqrt(1+x)-1}/x.

Solution:

If we put x = 0 in {sqrt(1+x)-1}/x we get 0/0 form of indetermination.

So we multiply by the conjugate surd {sqrt(1+x)+1} both numerator and denominator.

{sqrt(1+x)-1}/x = {sqrt(1+x)-1}{sqrt(1+x)-1}/ {x[(1+x)+1]}.

{sqrt(1+x)-1}/x = {(1+x)-1}/{x(sqrt(1+x)+1}.

{sqrt(1+x)-1}/x = x/{x(1+x)+1}.

{sqrt(1+x)-1}/x = 1/{sqrt(1+x)+1}.

Now we take the limit as x--> 0:

Lt x--> 0 {sqrt(1+x)-1}/x = Ltx-->0 1/{sqrt(1+x)+1} = 1/{sqrt(1+0)+1}.

Lt x--> 0 {sqrt(1+x)-1}/x = 1/2.