# Limits problem

1. Feb 17, 2015

### icecubebeast

1. The problem statement, all variables and given/known data
Find the limit analytically:
lim (x->4) [(x-3)^0.5 -1]/[x-4]
2. Relevant equations

3. The attempt at a solution
lim (x->4) [(x-3)^0.5 -1]/[x-4]
= lim (x->4) ([(x-3)^0.5 -1]/[x-4]) * [((x-3)^0.5 +1)/((x-3)^0.5 +1)]
= lim (x->4) [x - 3 -1]/[(x-4)(x-3)^0.5 +1)]
= [4 - 3 -1]/[(4-4)((4-3)^0.5 +1)]
= 0/ (1)
= 0

However my book says it's equal to 0.5
I don't know how to get to that answer.

Please don't say, "use L'hopital's rule" because this problem is assuming that you didn't learn derivatives, integrals, etc to solve the problem.

2. Feb 17, 2015

### DEvens

Have you done approximations of square root for numbers close to 1?

sqrt(1+d) approx. = 1 + d/2 for magnitude of d much less than 1

3. Feb 17, 2015

### Dick

The denominator doesn't come out to 1. It's also 0, but the x-3-1 in the numerator can cancel the x-4 in the denominator, right? THEN let x->4.

Last edited: Feb 17, 2015
4. Feb 17, 2015

### icecubebeast

Thank you so much! I didn't notice that those terms cancel.

5. Feb 17, 2015

### Ray Vickson

You substituted numbers too soon. You started with
$$f(x) \equiv \frac{\sqrt{x-3}-1}{x-4} = \frac{(\sqrt{x-3}-1)(\sqrt{x-3}+1)}{(x-4)(\sqrt{x-3}+1)} \\ = \frac{x-3-1}{(x-4)(\sqrt{x-3}+1)}.$$
Then you stopped simplifying, and that is where you ran into trouble. Why not go all the way? Continue to simplify, and write the last form as
$$\frac{1}{\sqrt{x-3}+1}$$
(That cancels the $(x-4)$ in the numerator and the denominator.) In other words, for all $x \neq 4$ we have
$$f(x) = \frac{\sqrt{x-3}-1}{x-4} = \frac{1}{\sqrt{x-3}+1} \equiv g(x)$$
The limit of $f$ is also the limit of $g$, and the latter is easy to get.