# Homework Help: Finding limit with two unknown numbers

1. Nov 5, 2011

### cataschok

1. The problem statement, all variables and given/known data

Find numbers a and b such that lim x$\rightarrow0$ [$\sqrt{}(ax+b)$ - 2] / x = 1

3. The attempt at a solution
so far, i've gotten to ax+b = (x+2)^2
... lim ax+b - lim (x+2)^2 = 0
...b - lim (x+2)^2 = 0
...b = 2^2
...b = 4

If i try to place b=4 back into the original equation, a will always end up as 0. I also cant find a way to extract a first.. So I'm stuck and I can't sleep at night.

This questions is supposed to "test our understandings of the material"... but I really can't understand much where each chapter is covered in a 1 hour lecture.

2. Nov 5, 2011

### Simon Bridge

It's supposed to test your understanding.... this is my personal bugbear so:

What does the concept of a "limit" mean?

$$\lim_{x \rightarrow 0} \frac{\sqrt{\left ( ax + b \right )} -2}{x} = 1$$

I can't tell if the -2 is inside the sqare root or not. Suspect outside from what you wrote after.

The first part looks good.
If you put b=4 into the original expression for the limit, you'll find the numerator and the denominator are both zero when x=0, which is what you need. But 0/0 is not always 1. So what is the other condition that ensures that the quotient is 1 in the limit?

Last edited: Nov 5, 2011
3. Nov 5, 2011

### cataschok

yes the equation that you wrote is correct

from what i got in the book, it says when both the numerator and denominator approaches zero, the best way to go at it would be to simplify the the quotient.

So I multiplied top and bottom with $\sqrt{}(ax+b)$ + 2

the numerator will eventually become ax+2 if i replace b with 4.
then if i take the limit of x->0, the result will be 2/4... so a = 1/2??

4. Nov 5, 2011

### HallsofIvy

First, since the denominator clearly goes to 0, in order that the limit exist at all, the numerator must also go to 0: setting x= 0 in the numerator, you must have $\sqrt{b}- 2= 0$. That tells you what b is. Now, what is a so that the limit is, in fact, 1?

(Your last sentence, "if i take the limit of x->0, the result will be 2/4... so a = 1/2??" makes no sense. What result? The "result" should either be 1 or involve a- setting those two equal should tell you what a must be.)

5. Nov 5, 2011

### Simon Bridge

I suspect the solution is in a temporary blind-spot, so it will probably not be guessed.
I don't want to just tell you ... so, I'll try another hint: I get a = 4, but how??[*]

In order for the limit to be one, the limit of the numerator must be the same as the limit of the denominator ... you got that part all right. But also, the numerator and the denominator must approach the limit at the same rate.

f=mx/nx has a limit at the origin of m/n, yes, because the x's cancel, but even without that because they approach 0 at different rates. The rate of approaching a limit is the understanding that was being tested.

Now - how would you find that out?

--------------------------------
What you did:
presumably hoping to get rid of the square-root [I tidied up your TeX]... but that just gives you:

$$\lim_{x\rightarrow0} \frac{ax+b-4}{x\sqrt{ax+b}}=1$$

I figure you reasoned like this:
putting b=4 lets you cancel the x's (bonus!):
$$\lim_{x\rightarrow0} \frac{a}{\sqrt{ax+4}}=1$$

in the limit, this becomes:
$$\frac{a}{\sqrt{4}} = 1 \Rightarrow a=2$$

... but if you put $y=\frac{\sqrt{2x+4}-2}{x}$ and plot (x,y), the curve approaches (0,0.5) instead of (0,1). So it cannot be right!

So what went wrong?

\sqrt{}(ax+b) +2

but it is clearer if it is
\sqrt{(ax+b)} + 2

LaTeX is so worth learning.

[*] I have been known to make silly arithmetic errors, sometimes on purpose, so don't take this as gospel.

Last edited: Nov 5, 2011
6. Nov 7, 2011

### cataschok

I'm approaching 0 solutions atm. lol but I had a last try

$\sqrt{(ax+b)} - 2$ = x
... $\sqrt{(ax+b)}$ = x + 2
ax + 4 = (x+2)^2
ax + 4 = x^2 + 4x + 4
ax = x ( x + 4 )
cancel out the x's

a = x+ 4
if lim x->0, a = 4

7. Nov 8, 2011

### Simon Bridge

well that worked :) just by direct cancellation.

8. Nov 8, 2011

### cataschok

To be honest, it'll help if you can explain to me a little more lol

9. Nov 8, 2011

### Simon Bridge

Do you have a tool for plotting functions on your computer?

If you plot that function, you'll find that it is defined everywhere except at the limit, where is just vanishes. You can get arbitrarily close to x=0, but not at it.

In an investigation, you can vary the value of a to discover how it changes the function.
This is a fast way to gain a feel for how these things work. That feeling is what guides the rest of us to the answers... rather than remembering the "right" method or plugging numbers into equations so it's worth getting.