Unknown Coefficients in a Limit

[Hammonds.C.Ryan]

I have been reviewing old math texts to prep for grad school. There is a problem that I cant seem to get beyond:

Find numbers a and b such that,
lim(x->0) [(sqrt(ax+b)-2)/x] =1.

I have tried using substitution with a parameter t=sqrt(ax+b), and also multiplying by the conjugate of the sqrt expression. But, I keep ending up unable to eliminate the division by 0. Also, I am attempting to do it without using L'Hopital's rule.

Any whiffs in the right direction would be appreciated.

 

[HallsofIvy]

First, since the denominator goes to 0, in order that the limit exist, the numerator must also go to 0. That is, \(\displaystyle \sqrt{a(0)+ b}- 2= \sqrt{b}- 2= 0\) so what must b equal? Now, do the usual with problems of this type: "rationalize" the numerator by multiplying both numerator and denominator by \(\displaystyle \sqrt{ax+b}+ 2: \frac{\sqrt{ax+b}- 2}{x}\frac{\sqrt{ax+b}+ 2}{\sqrt{ax+b}+2}= \frac{ax+ b- 4}{x(\sqrt{ax+b}+2)}\). Put in the value of b that you got before, take the limit as x goes to 0 and set it equal to 1. What must a equal?

 

[Hammonds.C.Ryan]

First, since the denominator goes to 0, in order that the limit exist, the numerator must also go to 0.

Thanks for the response. I completely agree with the method. However, the text from which I'm working does not address indeterminate forms until a later chapter. It's been a few years since I've done any sincere studying in math, so I'm trying to play along by not using knowledge the book has yet to teach me. That being said, I can't make the assumption about the numerator tending to 0, since it comes from a discussion on existence of limits in indeterminate form.

Any other ideas?

 

[HallsofIvy]

Knowing that if, in the fraction m/n, n goes to 0 and the m does not, the limit does not exist, has nothing to do with "indeterminant forms". You seem to be saying that you want a way to find this limit that require no knowledge of limits!

 

[JeffM]

Thanks for the response. I completely agree with the method. However, the text from which I'm working does not address indeterminate forms until a later chapter. It's been a few years since I've done any sincere studying in math, so I'm trying to play along by not using knowledge the book has yet to teach me. That being said, I can't make the assumption about the numerator tending to 0, since it comes from a discussion on existence of limits in indeterminate form.

Any other ideas?

Here is a way to think intuitively about the problem. The given function does not exist when x = 0. If a limit exists at all when x is near 0, then the function is not shooting off toward infinity. Right?

But that must mean that the function is NOT acting as though it is being divided by something close to 0. How can that be true? Why it must be because something is effectively canceling x. In short, we are looking for a function that at points close to x = 0 is equal to the given function but does not involve division by 0. Consequently, that other function will exist at x = 0. And what Halls is doing is systematically finding a function that (at points close to 0 but not equal to 0) is equal to the given function and also exists at x = 0.

My explanation makes no pretence of mathematical rigor, but what Halls did was as rigorous as you can imagine.

\(\displaystyle x = 0\ and\ \sqrt{ax + b} - 2 = 1 * x \implies \sqrt{(a * 0) + b} - 2 = 1 * 0 \implies \sqrt{b} = 2 \implies b = 4.\)

\(\displaystyle x \ne 0\ and\ \sqrt{ax + 4} - 2 = 1 * x \implies \left(\sqrt{ax + 4} - 2\right) * \left(\sqrt{ax + 4} + 2\right) = x\left(\sqrt{ax + 4} + 2\right) \implies (ax + 4) - 4 = x\left(\sqrt{ax + 4} + 2\right) \implies\)

\(\displaystyle ax = x\left(\sqrt{ax + 4} + 2\right) \implies a = \sqrt{ax + 4} + 2 \implies 1 = \dfrac{a}{\sqrt{ax + 4} + 2}.\)

But \(\displaystyle \sqrt{(a * 0) + 4} + 2 = \sqrt{4} + 2 = 2 + 2 = 4 \ne 0.\)

\(\displaystyle So\ f(x) = \dfrac{a}{\sqrt{ax + 4} + 2} \exists\ at\ x = 0.\)

\(\displaystyle f(0) = 1 \implies 1 = \dfrac{a}{\sqrt{(a * 0) + 4} + 2} = \dfrac{a}{4} \implies a = 4.\)

In other words,

\(\displaystyle x \ne 0 \implies \dfrac{\sqrt{4x + 4} - 2}{x} = f(x) = \dfrac{4}{\sqrt{4x + 4} + 2}.\)

That is easily checked by cross-multiplying.

And f(x) exists at x = 0 and has a limit of 1. So the other function has a limit of 1 at x = 0 even though it does not exist there. You can prove it if you want using the delta-epsilon method.

 

[dsk2]

Ryan, just checking, why dont you want to use L'Hopital's rule? Doesn't it make it considerably simple?

Using L'Hopital's rule

(1). You need have 0/0 form => The numerator has to be zero after substituting for x=0, thus sqrt(ax+b)-2=0. Hence b=4 straightaway.
(2). Since the (derivative of the numerator)/(derivative of denominator) at x=0 should be equal to 1. Knowing derivative of denominator is 1, derivative of numerator should be 1 at x=0. Thus,
\(\displaystyle \frac{a}{2 \sqrt(ax+b)}=1\) at \(\displaystyle x=0\). Thus \(\displaystyle a=2*\sqrt(4)=4\).

Cheers,
Sai.