# Limit of (sqrt(x) - 2)(x-4) as x approaches 4 from the left

## Homework Statement

$$\lim_{x\rightarrow 4^{-}}\frac{\sqrt{x}-2}{x-4}$$
Evaluate

## The Attempt at a Solution

Well, I used a calculator to substitute arbitrary values to see what it appears to approach.

input | output
2 .293
3 .268
3.5 .258
3.9 .252

So I concluded that the the limiting value is in fact .25.. and I was wrong. How do I evaluate this?

NOTE THAT WE ARE APPROACHING FROM THE LEFT!

Last edited:

Related Calculus and Beyond Homework Help News on Phys.org
Defennder
Homework Helper

Use L'Hospital's rule.

Thank you, I got it.

HallsofIvy
Homework Helper

A fairly standard way of handling a problem like that, and simpler than apply L'Hopital's rule is to rationalize the numerator. Multiplying both numerator and denominator by $\sqrt{x}+ 2$ you get
$$\frac{(\sqrt{x}-2)(\sqrt{x}+2)}{(x-4)(\sqrt{x}+2)}= \frac{x- 4}{(x-4)(\sqrt{x}+ 2)}$$
which, for x> 4, is
$$\frac{1}{\sqrt{x}+ 2}$$
Of course, the limit of that, as x goes to 2, is the same as the limit of the original problem.

A fairly standard way of handling a problem like that, and simpler than apply L'Hopital's rule is to rationalize the numerator. Multiplying both numerator and denominator by $\sqrt{x}+ 2$ you get
$$\frac{(\sqrt{x}-2)(\sqrt{x}+2)}{(x-4)(\sqrt{x}+2)}= \frac{x- 4}{(x-4)(\sqrt{x}+ 2)}$$
which, for x> 4, is
$$\frac{1}{\sqrt{x}+ 2}$$
Of course, the limit of that, as x goes to 2, is the same as the limit of the original problem.
Yeah, that is what I did. I'm not up to the rule yet. Thanks again halls of ivy.