One sided limit

JennyInTheSky

1. The problem statement, all variables and given/known data
lim[tex]_{x\rightarrow4^{-}}[/tex] [tex]\frac{\sqrt{x}-2}{x-4}[/tex]


2. Relevant equations
Typical methods used in solving one-sided limit.


3. The attempt at a solution
I plug in something a little bit bigger than four, like 4.0000001 into x, and I get [tex]\frac{something a little less than zero}{something a little less than zero}[/tex] to equal 1. But the answer is 1/4. How is this so?
1. The problem statement, all variables and given/known data



2. Relevant equations



3. The attempt at a solution

Gib Z

The idea with the limit of a quotient is not really about the values themselves, and them being close to zero, but what their limiting quotient is!

As you can see from as x-->0, 2x/x, the numerator and denominator will both be "something a little bigger than zero" as you put in smaller and smaller numbers, but we want to see the ratio of them in comparison to each other! Its quite obvious the limit is 2.

It's probably time you stopped using your calculator for the problems and started using some algebra =] Note the fact that [itex]x=\sqrt{x^2}[/itex] for x>0, which it is in this case, and use the difference of two squares formula [itex]a^2-b^2 = (a+b)(a-b)[/itex]. Then you can just cancel common factors and then sub x=4 straight in!

MiniST

Use L'hopital's rule since the top and bottom are either going to infinity or 0. Basically take the derivative of top and bottom and then take the limit. You can do this as many times as you like if the resulting expression satisfies the conditions.

JennyInTheSky

Thanks for your help, but you didn't have to be rude about it. I've had all of four days of high school calculus, so no, I'm not exactly comfortable with everything yet. I realize now that what I needed to use was algebra, but a simple nudge in that direction would have been sufficient, not wild accusations and assumptions. By the way, it's not quite so "obvious that the limit is 2." In fact, it's wrong. It's actually 1/4 because the limit as x approaches 4 from the negative side of [tex]\frac{1}{\sqrt{x}+2}[/tex] is 1/4.

JennyInTheSky

Thanks, but we have yet to learn derivatives!

Focus

You can divide two things that approach zero for example x/x=1 as x->0, because by the definition of a limit x is not zero. Once you gotten rid of the 0/0 s then you can just plug your limit in and get the answer.

HallsofIvy

I am puzzled as to your definition of "rude". You ask a question, someone makes valid suggestions, based entirely upon what YOU said you had done. You then accuse him of being rude, and making "wild accusations and assumptions". What wild accusations and assumptions did he make? He suggested that you not try using a calculator only after you had said "I plugged in a something a little bit bigger than 4" so it is certainly not a "wild accusation and assumption" that you were using a calculator. Nor is it a "wild accusation and assumption" that you did it incorrectly. If you put x= 4.0001, for example, you get
[tex]\frac{\sqrt{4.0001}- 2}{4.0001- 4}= .249998437..., nowhere near 1 but very close to 1/4.
Finally, Gib Z, in an effort to lead you to the correct idea, asked about 2x/x. The limit of THAT is 2 and that is what he was talking about.

If you consider that anyone who does not simply hand you the answer to be rude, then you are in the wrong forum.

JennyInTheSky

I did not want him to hand me the answer, and I'm glad he didn't. The way he responded just did not seem very cordial to me, but it's fine. Thank you very much to all who helped, I suppose that all this was was failure to do all of the algebra necessary in this problem.

BoundByAxioms

Just in case you were curious/for future reference: Using L'Hospital's rule for [tex]\frac{\sqrt{x}-2}{x-4}[/tex], you get: [tex]\frac{1}{2\sqrt{x}}[/tex]. Now just plug in 4, and you get 1/4.

HallsofIvy

Good, but "not cordial" does not equate to "rude".

The crucial point that Gib Z was trying to make is that [tex]x- 4= (\sqrt{x}-2)(\sqrt{x}+ 2)[/itex]. That's what you need.

JennyInTheSky

Indeed! Thanks