### Limits at infinity

Next, let's consider
$$\lim_{x\rightarrow \infty} {2x+3\over 5-x}$$
The hazard here is that $\infty$ is *not* a number that we can do
arithmetic with in the normal way. Don't even try it. So we *can't* really just ‘plug in’ $\infty$ to the expression to see what we
get.

On the other hand, what we really mean anyway is *not*
that $x$ ‘becomes infinite’ in some *mystical* sense, but rather
that it just ‘gets larger and larger’. In this context, the crucial
observation is that, as $x$ gets larger and larger, $1/x$ gets smaller
and smaller (going to $0$). Thus, just based on what we want this all
to mean,
$$\lim_{x\rightarrow \infty} {1\over x}=0$$
$$\lim_{x\rightarrow \infty} {1\over x^2}=0$$
$$\lim_{x\rightarrow \infty} {1\over x^3}=0$$
and so on.

This is the essential idea for evaluating simple kinds of limits as $x\rightarrow\infty$: rearrange the whole thing so that everything is expressed in terms of $1/x$ instead of $x$, and then realize that $$\lim_{x\rightarrow \infty}\;\;\;\;\hbox{ is the same as }\;\;\;\;\lim_{{1\over x}\rightarrow 0}$$

So, in the example above, divide numerator and denominator
both by *the largest power of $x$ appearing anywhere:*
$$\lim_{x\rightarrow \infty} {2x+3\over 5-x}=
\lim_{x\rightarrow \infty} {2+{3\over x} \over {5\over x}-1}=
\lim_{y\rightarrow 0} {2+3y\over 5y-1}={2+3\cdot 0\over 5\cdot
0-1}=-2$$

The point is that we called $1/x$ by a new name, ‘$y$’, and
rewrote the original limit as $x\rightarrow \infty$ as a limit as
$y\rightarrow 0$. Since $0$ *is* a genuine number that we can do
arithmetic with, this brought us back to ordinary everyday
arithmetic. Of course, it was necessary to rewrite the thing we were
taking the limit of in terms of $1/x$ (renamed ‘$y$’).

Notice that this is an example of a situation where we used the letter ‘$y$’ for something other than the name or value of the vertical coordinate.

#### Exercises

- Find $\lim _{x\rightarrow\infty} { x+1 \over x^2+3 }$.
- Find $\lim _{x\rightarrow\infty} { x^2+3 \over x+1 }$.
- Find $\lim _{x\rightarrow\infty} { x^2+3 \over 3x^2+x+1 }$.
- Find $\lim _{x\rightarrow\infty} { 1-x^2 \over 5x^2+x+1 }$.

#### Thread navigation

##### Calculus Refresher

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