# A question about limits of algebraic functions

I am a high school math teacher, and I recently put the following question on one of my assessments:

Find the numbers a and b such that $$\lim\limits_{x \to 0}\frac{\sqrt{ax+b}-4}{x}=1$$.

The concept I was testing was for students to realize that, since the limit exists but the denominator approaches 0, the numerator must also approach 0, i.e.

$$\lim\limits_{x \to 0}{\sqrt{ax+b}-4}=0$$.

Instead, some of my students made a different argument which seems a bit off to me. I don't want to tell them that their thinking is correct without being able to supply a justification for it. Note that, due to constraints in time and curriculum, we don't cover the definition of a limit (but I will gladly take an epsilon-delta proof if someone can supply one). I'm thinking, more likely, that there is a counter-example and if anyone can supply one I will be grateful. The student argument goes like this:

Consider $$\lim\limits_{x \to 0}\frac{x}{x}$$ We know that that limit has a value of 1. Note that the limit of the denominator is zero, i.e., $$\lim\limits_{x \to 0}{x} = 0$$

Since $$\lim\limits_{x \to 0}\frac{\sqrt{ax+b}-4}{x}$$ has the same value as $$\lim\limits_{x \to 0}\frac{x}{x}$$, and since the denominators of both expressions are equal and their limits are equal, we can say that the limit of the numerators must be equal, i.e. $$\lim\limits_{x \to 0}{\sqrt{ax+b}-4} = \lim\limits_{x \to 0}{x}$$.

From there, they arrive at the same result which I had intended by solving the limit on the right side of the equation. But something about this argument bothers me and I'm not certain what it is; perhaps it's because it is so specific in that the limit has to be going to zero, both denominators are just the identity function, and the original limit of the algebraic expressions is 1.

If I had made the original limit equal to 2, then could they have used their argument only changing the considered function from $$\frac{x}{x}$$ to $$\frac{2x}{x}$$? I feel like...maybe?

Anyway, can anyone provide a counter-example to this specific example of a limit approaching zero of a fractional expression whose denominators are identical, and where both approach zero, and the limit of the entire fractional expression is 1? Is there a counter-example, or can it be proven that this will always work? And then what about extrapolating it to when the limit of the original is an arbitrary constant k? And then when the denominators are unequal functions but both approach zero?

Thank you everyone for your help! I'm very big on rigorous explanations in my math classes (to the point where they can be made...I do an optional epsilon-delta lesson after school for those who are interested) and I'd like to either be able to tell my students that they are correct, or show them a counter-example. Thank you!

• +1 for "big on rigorous..." As we go up in math education, rigor is no longer optional. – Paramanand Singh Oct 22 '18 at 0:54

The argument by your student is correct (consider yourself lucky as a teacher)! The simplest approach here is the use of algebra of limits. We have $$\lim_{x\to 0} \sqrt{ax+b} - 4 =\lim_{x\to 0}\frac{\sqrt{ax+b} - 4}{x}\cdot x=1\cdot 0=0$$

The argument by the student can be expressed / justified as follows. Start with $$\lim_{x\to 0}\frac{\sqrt{ax+b}-4}{x}=1=\lim_{x\to 0}\frac{x}{x}$$ Now multiply both sides by $$\lim_{x\to 0}x$$ (which exists and is equal to $$0$$) and then using product rule of limits we get $$\lim_{x\to 0}\frac{\sqrt{ax+b}-4}{x}\cdot x=\lim_{x\to 0}\frac{x} {x} \cdot x$$ which leads to the desired conclusion.

You should note that the approach by your student is essentially the same as my approach given in first half of this answer. But I think you need to explain the details because there may be a chance that the student is not aware of it.

The typical approach via contradiction is more popular (if the numerator does not go to $$0$$ but denominator does then fraction can't tend to a finite limit), but I find the direct use of algebra of limits simpler.

• That makes sense! That's what I was missing! The product of the limits theorem can be used to rewrite the product of limits as a limit of products! And since the denominators are equivalent functions, they wind up cancelling with the additional function multiplied in the limit! That makes sense! Thank you! – analysischallenged Oct 22 '18 at 0:48
• @analysischallenged: I am glad you liked my answer. You should always grab any opportunity to highlight the importance of algebra of limits. They are simple and powerful, but sadly ignored by many students in favor of L'Hospital's and Taylor while evaluating limits. – Paramanand Singh Oct 22 '18 at 0:51
• Absolutely! I didn't realize this myself, but rest assured I will make it a point to demonstrate this in class this week! Thank you! – analysischallenged Oct 22 '18 at 0:53

I don't have one off the cuff but I'm fairly certain there is a counter example. Since the limit of the denominator is 0 we don't have

$$\lim_{x \to 0}\frac{x}{x}=\frac{\lim_{x \to 0}x}{\lim_{x \to 0}x} = \frac{0}{0}$$

so we can't treat them algebraically. The observation your students should make is that for this limit to be 1 the numerator needs to go to $$0$$ exactly as fast as the denominator does. Then, we can say we want $$\lim_{x \to 0}\sqrt{ax+b}-4-x=0$$

• Even if there isn't a counter example it doesn't work for the reason you stated just for the record. – Aaron Zolotor Oct 21 '18 at 18:05