Effective Trick for Limits of Rational Functions

In calculus, people are interested in calculating limits such as the following:

$$ \lim_{x \to \infty} \frac{3x^3 + 5}{2x^3 \; – \; 1,000,000\,x^2} $$

To calculate this, we need to know what the fraction

$$ \frac{3x^3 + 5}{2x^3 \; – \; 1,000,000\,x^2} $$

looks like as \( x \) gets larger and larger as a positive number.

In today’s blog, I’ll explain why the answer to the above is \(1.5\).

It seems that the above answer is unlikely. For instance, if I set \(x = 1000\) and calculated the above fraction, I get:

$$ \frac{3 \cdot 1000^3 + 5}{2 \cdot 1000^3 \; – \; 1,000,000 \cdot 1000^2} $$ $$ =\frac{600,000,000}{199,600,000,000} $$ $$ = -0.003006012\cdots $$

This is very different from \(1.5\). But appearances are deceiving.

Let’s take a closer look at that fraction. Divide both the numerator and denominator by \( x^3 \):

$$ \frac{3x^3 + 5}{2x^3 \; – \; 1,000,000 \, x^2} $$ $$ = \frac{(3x^3 + 5)/x^3}{(2x^3 \; – \; 1,000,000 \, x^2)/x^3} $$ $$ = \frac{3x^3 /x^3+ 5/x^3}{2x^3/x^3 \; – \; 1,000,000 \, x^2/x^3} $$ $$ = \frac{3 + 5/x^3}{2 \; – \; 1,000,000/x} $$

In summary,

$$ \frac{3x^3 + 5}{2x^3 \; – \; 1,000,000 \, x^2} = \frac{3 + 5/x^3}{2 \; – \; 1,000,000/x} $$

Why is this significant? Look at \( 5/x^3 \) and \(1,000,000/x\). What both of these have in common is that they become smaller and smaller as \( x \) gets larger. In fact, they both approach \( 0 \).

So as \( x \) gets larger, the fraction

$$ \frac{3 + 5/x^3}{2 \; – \; 1,000,000/x} $$

gets closer and closer to

$$ \frac{3 + 0}{2 \, – \, 0} = 1.5 $$

And that is our answer. In summary:

$$ \lim_{x \to \infty} \frac{3x^3 + 5}{2x^3 \; – \; 1,000,000 \, x^2} $$ $$ = \lim_{x \to \infty} \frac{3 + 5/x^3}{2 \; – \; 1,000,000/x} $$ $$ = \frac{3 + 0}{2 \, – \, 0} \\ \\ = 1.5 $$