Is the Method Used to Solve for a and b in this Calculus Problem Valid?

[samtouchdown]

This problem's solution has been widely debated for its validity. Though it gives the correct answer, many argue that it is invalid. Please give comments on why this method works and if it is valid or not. If possible, please provide your own solution. The problem is in the attachment.

 

[Chirag B]

Yes, if a limit takes on the indeterminate form 0/0, it can take on any value. In actuality, one could also solve for the limit using l'Hospital's Rule.

Differentiating the top and bottom leaves us with this statement: \lim_{x \rightarrow 0}\frac{\frac{a}{2\sqrt{ax + b}}}{1} = 1, which is the same as: \lim_{x \rightarrow 0}\frac{a}{2\sqrt{ax + b}} = 1.

Now, if we plug in x = 0, we get: a = 2\sqrt{b}. If we square both sides, we get a^{2}=4b. If we plug this back into the original equation (without using l'Hospital's Rule), we will find that a = 4. Then, using the above statement, it follows that b = 4 also.

 

[micromass]

That OP made me pull out my hair. You can't do something like

\lim_{x\rightarrow 0}{f(x)}=x

x is a variable in the limit. It's a dummy variable. You can't just take it out and put it on the other side. Doing that shows a fundamental lack of understanding limits! Please don't ever do that!

 

[Chirag B]

That OP made me pull out my hair. You can't do something like

\lim_{x\rightarrow 0}{f(x)}=x

x is a variable in the limit. It's a dummy variable. You can't just take it out and put it on the other side. Doing that shows a fundamental lack of understanding limits! Please don't ever do that!

I agree. The only way to solve this problem was either by (a) splitting the limit into a quotient with the top being its own limit and bottom being its own limit and then multiplying by \lim_{x\rightarrow 0}(x) on both sides, or (b) using l'Hospital's Rule as I mentioned above. Good insight.

 

[micromass]

splitting the limit into a quotient with the top being its own limit and bottom being its own limit and then multiplying by lim_{x→0}(x) on both sides

No, you can't do that. That would yield a 0/0 situation which is not allowed. Splitting the limit into a quotient is not allowed here!

 

[Chirag B]

No, you can't do that. That would yield a 0/0 situation which is not allowed. Splitting the limit into a quotient is not allowed here!

Ah, well then I was wrong. But as far as I know, l'Hospital's Rule still applies. You were angry at the poster's making the mistake of taking the variable x outside of the limit, I presume?

 

[micromass]

Ah, well then I was wrong. But as far as I know, l'Hospital's Rule still applies. You were angry at the poster's making the mistake of taking the variable x outside of the limit, I presume?

I wasn't angry just a bit disappointed.

Yes, L'Hopitals rule should apply.

 

[Chirag B]

I wasn't angry just a bit disappointed.

Yes, L'Hopitals rule should apply.

Makes sense.

 

[jgens]

There is no need for L'Hopitals' Rule here! In fact, we have,
\lim_{x \to 0}\frac{\sqrt{ax+b}-2}{x} = \lim_{x \to 0}\frac{ax+b-4}{x(\sqrt{ax+b}+2)} = 1
But since the denominator goes to 0, this forces the numerator to go to 0 if the limit is to exist. Therefore, lim_x→0ax+b-4 = 0, which forces b = 4. Then we have
\lim_{x \to 0}\frac{ax}{x(\sqrt{ax+4}+2)} = \lim_{x \to 0}\frac{a}{\sqrt{ax+4}+2} = 1
From here it is pretty obvious that this forces a = 4.

 

[Curious3141]

There are 2 obvious ways to do this: L'Hopital's (already shown) and Taylor series (my favourite).

\lim_{x -&gt; 0} \frac{\sqrt{4x+4} - 2}{x} = 2\lim_{x -&gt; 0} \frac{\sqrt{x+1} - 1}{x}<br /> <br /> = 2\lim_{x -&gt; 0} \frac{1 + \frac{x}{2} - 1}{x} = 1

 

[SammyS]

This problem's solution has been widely debated for its validity. Though it gives the correct answer, many argue that it is invalid. Please give comments on why this method works and if it is valid or not. If possible, please provide your own solution. The problem is in the attachment.

micromass has already pointed out that it contains non-sense statements.

It's not particularly difficult to come up with a method which seems to work, when it contains statements that profoundly abuse accepted notation.

 

[Curious3141]

There are 2 obvious ways to do this: L'Hopital's (already shown) and Taylor series (my favourite).

\lim_{x -&gt; 0} \frac{\sqrt{4x+4} - 2}{x} = 2\lim_{x -&gt; 0} \frac{\sqrt{x+1} - 1}{x}<br /> <br /> = 2\lim_{x -&gt; 0} \frac{1 + \frac{x}{2} - 1}{x} = 1

Sorry, I had only taken a cursory reading of the problem earlier.

I understand that the requirement is to determine a and b. I think the determination of b was done correctly in the OP's solution. So b = 4.

The steps to determine a were done wrongly because in a limit, one cannot bring the x over to the RHS like that.

Instead of this, LH or Taylor/Binomial series are perfectly acceptable.

Sticking with the Taylor/Binomial approach:

LHS = \lim_{x -&gt; 0} \frac{\sqrt{ax+4} - 2}{x}<br /> <br /> = \lim_{x -&gt; 0} \frac{2\sqrt{1+\frac{ax}{4}} - 2}{x}<br /> <br /> = \lim_{x -&gt; 0} \frac{2(1 + \frac{ax}{8}) - 2}{x}<br /> <br /> = \frac{a}{4}

Equating that to the RHS of 1,

a = 4

 

[Chirag B]

Sorry, I had only taken a cursory reading of the problem earlier.

I understand that the requirement is to determine a and b. I think the determination of b was done correctly in the OP's solution. So b = 4.

The steps to determine a were done wrongly because in a limit, one cannot bring the x over to the RHS like that.

Instead of this, LH or Taylor/Binomial series are perfectly acceptable.

Sticking with the Taylor/Binomial approach:

LHS = \lim_{x -&gt; 0} \frac{\sqrt{ax+4} - 2}{x}<br /> <br /> = \lim_{x -&gt; 0} \frac{2\sqrt{1+\frac{ax}{4}} - 2}{x}<br /> <br /> = \lim_{x -&gt; 0} \frac{2(1 + \frac{ax}{8}) - 2}{x}<br /> <br /> = \frac{a}{4}

Equating that to the RHS of 1,

a = 4

That's a nice way to do it...interesting!