Question

Question #adcfb

Answer 1

`= 1/(2 sqrt 3)`

Explanation:

`lim_(x->0) (sqrt(3+x)-sqrt3)/x`

`= lim_(x->0) (sqrt(3)(1+x/3)^(1/2)-sqrt3)/x`

By Binomial Expansion, with `0 < abs x " << " 1`

`= lim_(x->0) (sqrt(3)(1+x/6 + O(x^2))-sqrt3)/x`

`= lim_(x->0) sqrt(3)/6 + O(x)`

`= sqrt(3/36) = 1/(2 sqrt 3)`

Answer 2

(Alternate approach, using L'Hospital's Rule).
`sqrt3/6`

Explanation:

L'Hospital's Rule: if we have an indeterminate form `0/0`, we need differentiate the numerator and the denominator and then take the limit.
Given is the problem
`lim_(x->0) (sqrt(3+x)-sqrt3)/x`
We observe that when we take the limit and insert `x=0`, the expression becomes `0/0`. Applying the above stated rule we get
`lim_(x->0) (d/dx(sqrt(3+x)-sqrt3))/(d/dxx)`, rewriting numerator in the exponent form
`lim_(x->0) (d/dx((3+x)^(1/2)-sqrt3))/(d/dxx)`
`lim_(x->0) (1/2(3+x)^(-1/2)xx1)/1`
Now taking the limits
`1/2(3+0)^(-1/2)`
`=>1/(2sqrt3)`
Rationalizing the denominator we get
`sqrt3/6`

Answer 3

"Rationalize" the numerator.

Explanation:

`((sqrt(3+x)-sqrt3))/x * ((sqrt(3+x)+sqrt3))/((sqrt(3+x)+sqrt3)) = ((3+x)-3)/(x(sqrt(3+x)+sqrt3))`

`= x/(x(sqrt(3+x)+sqrt3)) = 1/(sqrt(3+x)+sqrt3)`

`lim_(xrarr0)1/(sqrt(3+x)+sqrt3) = 1/(sqrt(3+0)+sqrt3) = 1/(2sqrt3)`