Question

How do you evaluate `lim_(x->0) (sqrt(4+x)-2)/(3x)`?

Answer 1

Given: `lim_(x to 0)(sqrt(4+x)-2)/(3x)`

Because the limit yields the indeterminant form `0/0`, one should use L'Hôpital's Rule. But, because the topic is "Determining Limits Algebraically", I shall assume that the student has not, yet, learned L'Hôpital's Rule and refrain from using it.

Multiply the expression by 1 in the form of `(sqrt(4+x)+2)/(sqrt(4+x)+2)`:

`lim_(x to 0)(sqrt(4+x)-2)/(3x)(sqrt(4+x)+2)/(sqrt(4+x)+2)`

The numerator becomes the difference of two squares:

`lim_(x to 0)((sqrt(4+x))^2-(2)^2)/((3x)(sqrt(4+x)+2))`

Expand the squares:

`lim_(x to 0)(4+x-4)/((3x)(sqrt(4+x)+2))`

Simplify the numerator:

`lim_(x to 0)x/((3x)(sqrt(4+x)+2))`

`x/x` becomes 1:

`lim_(x to 0)1/(3(sqrt(4+x)+2))`

Now, we may evaluate at `x = 0`:

`1/(3(sqrt(4)+2)) =1/12`

This limit is the same as the original expression:

`lim_(x to 0)(sqrt(4+x)-2)/(3x) = 1/12`

Answer 2

`lim_(x->0) (sqrt(4+x)-2)/(3x) = 1/12`

Explanation:

Apply L'Hospital's Rule :

`lim_(x->0) (sqrt(4+x)-2)/(3x)`

`= lim_(x->0) ((d((4+x)^(1/2)-2))/(dx))/((d(3x))/(dx))`

`=lim_(x->0) (1/2(4+x)^(-1/2))/(3)`

`=lim_(x->0) 1/(6sqrt(4+x))`

`=1/(6sqrt(4))`

`=1/12`