Question

How do you find `lim (sqrt(x+1)+1)/(sqrt(x+1)-1)` as `x->0^+` using l'Hospital's Rule or otherwise?

Answer 1

This is not an indeterminate form. See below.

Explanation:

As `xrarr0`, the numerator goes to `2`. Therefore, the form is not indeterminate and it would be an error to use l'Hospital's Rule.

As `xrarr0`, the denominator goes to `0`, so the limit from the right is either `oo` or `-oo`.

To determine which, consider an `x`, just a bit greater than `0`.

Then `x+1` is a bit more than `1`, and `sqrt(x+1)` is a bit greater than `1`. So the denominator is a positive number near `0`.

We can write the form of the limit as `2/0^+` which tells us that, as `xrarr0^+`, the quotient increases without bound.

We write

`lim_(xrarr0^+) (sqrt(x+1)+1)/(sqrt(x+1)-1) = oo`

Answer 2

There is no limit.

Explanation:

Generally:

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

`= lim_(xto 0) ((1 + 1/2x + ...) +1)/((1 + 1/2 x + ...)-1)`

`= lim_(xto 0) ((2 + 1/2x + ...) )/((1/2 x + ...))`

For small positive `x`, this is always going to be positive. For small negative `x`, this is always going to be negative.

So there is no limit :(

We have a 2-sided limit.