There is one more kind of limit that I will cover now briefly. The limit concept is the same; it's just another way that we can approach the limit. Consider the function:

xThis function is graphed here in figure 2.3-1. Notice that as^{2}- 1 f(x) =~~eq. 2.3-1 x~~^{2}+ 1

f(1) = 0 tab. 2.3-1 f(2) = 0.6 f(5) = 0.9231 f(10) = 0.9802 f(100) = 0.9998 f(1000) = 0.999998We have a feeling that this function can get as close to

So how close does `f(x)` get to `1`? We find that by taking
the absolute value of the difference. After you put both terms over a
common denominator, you get:

2 |f(x) - 1| =The denominator clearly gets big without limit as~~eq. 2.3-2 x~~^{2}+ 1

æ 2 ö x ³ sqrtç - - 1 ÷ eq. 2.3-3 è e øand you'll be guaranteed it." (noting that

This is a situation where `f(x)` approaches a limit as `x`
grows without limit, or, in the popular nomenclature, as `x` goes to
`¥` (which is the symbol for infinity). So we say:

lim f(x) = L eq. 2.3-4 xto mean that for any~~> ¥~~

|f(x) - L| £ e eq. 2.3-4whenever

Here is a useful trick for taking a limit as `x` goes to infinity.
Supose you had, for example

x (1 - x) limIf you substitute~~eq. 2.3-5 x~~~~> ¥ (2 + x)(3 + x)~~

1 x =throughout and then take the limit as~~eq. 2.3-6 u~~

1 æ 1 öAt first, this looks pretty nasty, but if you multiply numerator and denominator by~~ç 1 -~~~~÷ u è u ø lim~~~~eq. 2.3-7 u~~~~> 0 æ 1 ö æ 1 ö ç 2 +~~~~÷ ç 3 +~~~~÷ è u ø è u ø~~

(1) (u - 1) limSince this thing is already factored for you, you can apply the product rule for limits (remember,~~eq. 2.3-8 u~~~~> 0 (2u + 1)(3u + 1)~~

1 limand~~eq. 2.3-9a u~~~~> 0 2u + 1~~

u - 1 limThese two are each pretty easy. If you put zero in for~~eq. 2.3-9b u~~~~> 0 3u + 1~~

This method *always* works for finding limits as something goes
to infinity. Substitute `1/u` for that something everywhere
it occurs, then take the limit as `u` goes to zero using whatever
methods are in your arsenal for taking such limits. Often you
will have to simplify first by multiplying numerator and denominator
by some power of `u` in order to clear the little fractions.
But you learned how to do that in algebra.

You could even do a `delta-epsilon` proof of the limit as `u`
goes to zero. Since `x` goes to infinity as `u` goes to zero,
you can argue that showing that finding a `d` close enough to
zero to meet the `e` contract for the `u`-version
is the the problem
same as finding an `x` large enough to meet the same `e`
contract for the `x`-version of the problem. In fact, whenever
` |u| £ d`` |x| ³ 1/d`

1) See if you can come up with the definition of what it means for a function,
`f(x)`, to have a limit as `x` goes to `-¥`
(that is, as `x` gets negative without limit). Apply that definition
to the function in 2.3-1, and show that it, indeed, approaches a limit
of `1` as `x` goes to `-¥`.

2) Find and prove the limit of

3x + 3 f(x) =as~~x - 1~~

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