Course:MATH110/Archive/2010-2011/003/Notes/Limits at infinity

Limits at infinity

Let's look at the sequence

$\displaystyle {a_{0}=3}\quad a_{1}={\frac {11}{5}}=2.2\quad a_{2}={\frac {19}{9}}\approx 2.111\quad a_{3}={\frac {27}{13}}\approx 2.077\quad a_{4}={\frac {35}{17}}\approx 2.059\quad a_{5}={\frac {43}{21}}\approx 2.048\quad \ldots$

We see that at each step, the numerator adds 8 and the denominator adds 4. This means we can rewrite this as

$a_{0}={\frac {3}{1}}\quad a_{1}={\frac {3+8}{1+4}}\quad a_{2}={\frac {3+2\cdot 8}{1+2\cdot 4}}\quad a_{3}={\frac {3+3\cdot 8}{1+3\cdot 4}}\quad a_{4}={\frac {3+4\cdot 8}{1+4\cdot 4}}\quad a_{5}={\frac {3+5\cdot 8}{1+5\cdot 4}}\quad \ldots$

Now that we see the pattern, we can deduce that if we want this pattern to persist, then the sequence has to have general formula:

$a_{n}={\frac {3+8n}{1+4n}}$

Since this course is about real-valued functions, it is worth noting that we can create the function $f$ defined by:

$f(x)={\frac {3+8x}{1+4x}}$

Which is defined for every real value $x$ except when $x=-1/4$. Now, the values of the sequence are simply the values of this function at all positive integers, or $a_{n}=f(n)$.

Whether we look at the sequence or the function doesn't really matter. It is not difficult to convince yourself that the behaviour of the sequence is dictated by the behaviour of the function. From the numerical values we have, it seems that the larger $x$ gets, the closer to 2 the values of $f(x)$ becomes. Can we make sure of this?

To study this question, we would like to see how far from $2$ the values of $f(x)$ are. To do this, we simply look at their difference and since we're not interested in knowing if this amount is positive or negative, we take its absolute value. This can be written as $|f(x)-2|$. (Note this is $f(x)-2$ inside absolute values, there's no 21).

Let's compute this amount precisely in this case. We have that

$\left|f(x)-2\right|=\left|{\frac {3+8x}{1+4x}}-2\right|=\left|{\frac {3+8x}{1+4x}}-{\frac {2+8x}{1+4x}}\right|=\left|{\frac {3+8x-2-8x}{1+4x}}\right|=\left|{\frac {1}{1+4x}}\right|$

Now, the absolute values are a little bit annoying, can we get rid of them? Well, clearly if $x>-1/4$ then $1+4x>0$. Since we're interested in looking at this function as the values of $x$ grow very large, it makes sense to say that we actually will not need the absolute values since this means the quantity inside is already positive. Hence, our new question is now how little does

$\left|f(x)-2\right|={\frac {1}{1+4x}}\qquad {\text{(if }}x>-0.25{\text{)}}$

become as $x$ grows larger? We'll say that $2$ is the limit, if we can actually make this as little as we want, granted that we let the values of $x$ be large enough. Let's try to do this. Let us fix a very small positive quantity that we'll denote by $\epsilon$ (epsilon). You get to chose how small you want this to be. For example, it could be 0.0000001, but everyone can chose a different value. What matters, is that however small we chose this value, we want to see that actually all the values of the function $f$ will be that close to $2$ once $x$ becomes large enough.

So, let's try to see if we can do this and actually tell you how large $x$ has to be for the function to be $\epsilon$-close to the value $2$. Again, what are we looking for? We want to know what values of $x$ make this true:

$\left|f(x)-2\right|<\epsilon$

How do we do this? Well, we just have a look! We have now a precise formula for the left-hand side of that equation, let's see what it tells us.

Since

$\left|f(x)-2\right|={\frac {1}{1+4x}}$

we can write that for

$\left|f(x)-2\right|<\epsilon$

to be true, we need that

${\frac {1}{1+4x}}<\epsilon$

or in other words that

${\frac {1}{\epsilon }}<1+4x$

which is equivalent to say that

$x>{\frac {1}{4}}\left({\frac {1}{\epsilon }}-1\right)$

Which makes a lot of sense. What we see here is telling us that as soon as $x$ is larger than this complicated value which depends of the value $\epsilon$ we chose, then the function is $\epsilon$-close to $2$. When you have a close look at it, it does make sense, since if $\epsilon$ is very small, then $1/\epsilon$ should be very big. Then this says that $x$ has to be larger than a quarter of that big number (to which 1 was substracted).

Let's test this. What does this tell us if we want to see when is the function 0.000001-close to $2$? Well, since

$0.000001={\frac {1}{1'000'000}}$

we have that

${\frac {1}{0.000001}}=1'000'000$

and hence that if $\epsilon =0.000001$, then

${\frac {1}{4}}\left({\frac {1}{\epsilon }}-1\right)={\frac {1}{4}}(1'000'000-1)\approx 250'000$

In other words, the values of $f(x)$ will be less than $2.000001$ for all values of $x$ larger than $250'000$.

The real point, is that we found a way to show that however small we take the value of $\epsilon$ to be, we can say how large $x$ has to be for ALL the further values of $f(x)$ to be that close to $2$. This is what allows us to say that the limit at infinity of this function is actually 2.

In other words, this constitute a proof that

$\lim _{x\rightarrow +\infty }{\frac {3+8x}{1+4x}}=2$

The definition of a limit at infinity

Let's have a look again at our definition:

We say that a function $f$ admits the number $L$ as its limit at infinity if the values of $f(x)$ gets arbitrarily close to $L$ as the values of $x$ gets larger.

More precisely, we say that a function $f$ admits the number $L$ as its limit at infinity if for any choice of $\epsilon >0$, we can find a number $N_{\epsilon }$ such that

$\left|f(x)-L\right|<\epsilon$

for all the values $x>N_{\epsilon }$.

Note that in our example, we actually found a precise formula to find the value $N_{\epsilon }$ which says that

$N_{\epsilon }={\frac {1}{4}}\left({\frac {1}{\epsilon }}-1\right)$

Questions to ponder now

• So, what would it mean for a function not to have a limit at infinity? Can you find examples of this?
• How do we find the value $L$ in general? In the example above, we kinda guessed...