Question

Evaluate the limit of the indeterminate quotient?

Answer 1

`lim_(x->0) (sqrt(7-x)-sqrt(7+x))/x = -1/sqrt(7)`

Explanation:

Evaluate the limit:

`lim_(x->0) (sqrt(7-x)-sqrt(7+x))/x`

Rationalize the numerator of the function using the identity: `(a+b)(a-b) = (a^2-b^2)`:

`(sqrt(7-x)-sqrt(7+x))/x = ((sqrt(7-x)-sqrt(7+x))/x)((sqrt(7-x)+sqrt(7+x))/(sqrt(7-x)+sqrt(7+x))) = (7-x-7-x)/(x(sqrt(7-x)+sqrt(7+x))) = -(2x)/(x(sqrt(7-x)+sqrt(7+x)) ) = -2/(sqrt(7-x)+sqrt(7+x))`

Now the limit is determinate:

`lim_(x->0) (sqrt(7-x)-sqrt(7+x))/x = lim_(x->0) -2/(sqrt(7-x)+sqrt(7+x)) = -1/sqrt(7)`

Answer 2

`-1/sqrt7`

Explanation:

Here is another way of solving it. Slightly messy.

L'hopital's Rule:

`lim_(x->0)(f(x))/(g(x))=lim_(x->0)(f^'(x))/(g^'(x))` only if the expression is indeterminate .

`f^'(x)` is simply the derivative of `f(x)` with respect to `x`.

Now via L'hopital's Rule (differentiate the numerator and the denominator separately), `lim_(x->0)=(sqrt(7-x)-sqrt(7+x))/x=lim_(x->0)(-1/2(7-x)^(-1/2)-1/2(7+x)^(-1/2))/1=-1/2(1/(sqrt(7-x))+1/(sqrt(7+x)))`

Let `x=0` and you get `-1/sqrt7`