Analytically Finding Limits: (x-3)^0.5 -1 / x-4

[icecubebeast]

Homework Statement

Find the limit analytically:
lim (x->4) [(x-3)^0.5 -1]/[x-4]

Homework Equations

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.

 

[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

 

[Dick]

Homework Statement

Find the limit analytically:
lim (x->4) [(x-3)^0.5 -1]/[x-4]

Homework Equations

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.

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.

 

[icecubebeast]

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.

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

 

[Ray Vickson]

Homework Statement

Find the limit analytically:
lim (x->4) [(x-3)^0.5 -1]/[x-4]

Homework Equations

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.

You substituted numbers too soon. You started with
$$f(x) \equiv \frac{\sqrt{x-3}-1}{x-4} = \frac{(\sqrt{x-3}-1)(\sqrt{x-3}+1)}{(x-4)(\sqrt{x-3}+1)} \\<br /> = \frac{x-3-1}{(x-4)(\sqrt{x-3}+1)}.$$
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.