Dear Uncle Colin

I have been asked to describe how $y = \frac{3x^2-1}{3x+2}$ behaves as $x$ goes to infinity. My first answers, “$y$ goes to infinity” and “$y$ approaches $x$”, were both wrong. Any ideas?

- Both Options Reasonable, Erroneous Limits

Hi, BOREL, and thanks for your message!

My first answer was $y$ approaches $x$ as well - clearly, as $x$ gets large, the only terms with any real impact are the leading terms on the top and the bottom, so you’re left with $\frac{3x^2}{3x}$, which is $x$. However, there’s more to it - you have a quadratic expression divided by a linear one, so you should end up with a linear answer - which is to say, something of the form $ax+b$.


A simple way to work this through is to work through the division process. You might prefer long division; I quite like matching coefficients; but generally, I recommend a grid method as the most coherent way to divide polynomials.

Whichever approach you take should give $\frac{3x^2-1}{3x+2} = x-\frac{2}{3} - \frac{\frac{4}{3}}{3x+2}$ - and the oblique asymptote is $y=x-\frac{2}{3}$.


I had a lecturer from Germany who routinely used the phrase “If we adopt the Ansatz…” - I’m not sure there’s an English translation, but it’s roughly “template solution”. You know what the answer ought to look like, and just need to fill in the details.

In this case, we’re looking at $\frac{3x^2-1}{3x+2}$, which ought to be close to $ax+b$ as $x$ gets large. That is to say, $\frac{3x^2-1}{3x+2} - (ax + b)$ should go to zero as $x$ increases.

If we turn that into a single fraction, we get $\frac{3x^2 - 1 - (ax+b)(3x+2)}{3x+2} = \frac{3x^2 - 1 - 3ax^2 - 3bx - 2ax -2b}{3x+2}$.

Matching the coefficients of $x^2$ and $x$, $3a-3=0$ and $-3b-2a=0$. Almost trivially, $a=1$ and $b=-\frac{2}{3}$.

An engineer’s approach

If $x=100$, $\frac{3x^2-1}{3x+2} = \frac{29,999}{302} \approx 99.3344…$. That looks like it’s about two-thirds less than $x$ to me! This approach is not recommended for sentient humans, though.

Hope that helps!

- Uncle Colin