On "Proving' Infinite Limits, And A Counterexample

Question

The textbook I am using has come up with this method to prove why in infinite limits, the term with the highest degree will dominate:

It then goes on to say that at infinity, 1000/3x and all the other terms will be virtually 0. Therefore, the function will approach 1 (since it it lim x->infinity (1-0+0-0).

However, my problem with this, is for instance, take a function x+2/x-1. On the logic of the above, when we break it up, it is (x+2)*(1/x-1). At infinity, 1/x-1=0. So therefore, the limit is 0. But the limit of x+2/x-2 is actually 1.

We see this again we we take a function sin(x)*1/x. Again, at infinity, the limit should be 0, but it's not.

So what can explain this contradiction? Why is the above proof still right in these situations despite the examples below? How can we know that the proof above is universally true? And why is my logic wrong? Why is is x+2/x-1 not equals to 0?

Answer 1

Applying the same method to $\frac{x+2}{x-1}$, what we get is\begin{align}\lim_{x\to\infty}\frac{x+2}{x-1}&=\lim_{x\to\infty}\frac{(x-1)+3}{x-1}\\&=\lim_{x\to\infty}1+\frac3{x-1}\\&=1.\end{align}

And the limit $\lim_{x\to\infty}\frac{\sin x}x$ is $0$.

Answer 2

[T]ake a function $\frac{x+2}{x-1}$. On the logic of the above, when we break it up, it is $(x+2)\frac{1}{x-1}$. At infinity, $\frac{1}{x-1}$ [tends to] $0$. So therefore, the limit is 0. But the limit of $\frac{x+2}{x-1}$ is actually $1$.

It's very good that you're noticing what seems to be a contradiction here. The flaw in your reasoning is that the limit laws can only be used when the constituent limits exist. For instance, the product law for limits is

If $\lim_{x\to\infty}f(x) = L$ and $\lim_{x\to\infty} g(x) = M$, then $\lim_{x\to\infty} f(x)g(x) = LM$.

You're trying to apply this law to the situation $f(x) = x+2$ and $g(x) = \frac{1}{x+1}$. Yes, $\lim_{x\to\infty} g(x) = 0$ and “anything times zero is zero.” However, $\lim_{x\to\infty} f(x)$ is not a “thing”—it doesn't exist. So we cannot apply the product law.

Students get confused because we say $\lim_{x\to\infty}f(x) =\infty$, and because there's an equals sign and a symbol on the other side, it sure looks like the limit “is” something. But $\lim_{x\to\infty}f(x) =\infty$ is shorthand for an infinite limit. There is some arithmetic than can be done with infinity, but $\infty\cdot0$ is undefined.

We see this again we we take a function $\sin(x)\frac{1}{x}$. Again, at infinity, the limit should be 0, but it's not.

Again, the first factor has no limit as $x\to\infty$, so the product law for limits doesn't apply.

Answer 3

There is no contradiction at all.

The rule says that in a sum of terms you can neglect the terms that tend to zero (unless all the terms tend to zero).

The rule never said that this holds for a product.

For instance, consider $x+\dfrac1x$ vs. $x\cdot\dfrac1x$.

For increasing powers of $10$, we have

$$x+\frac1x=2,10.1,100.01,1000.001,10000.0001\cdots$$

which clearly tends to $x$, vs.

$$x\cdot\frac 1x=1,1,1,1,1,\cdots$$

which is intermediate between $x$ and $\dfrac1x$ and doesn't vanish.

For these reasons, you can write statements such as

$$\lim_{x\to\infty}\frac{2x+3}{3x-5}=\lim_{x\to\infty}\frac{2x}{3x}=\frac25.$$