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Level 8 - NCEA Level 3

Evaluating Limits


In this chapter, we ask how to decide whether or not a limit exists and then, how to evaluate it. We are thinking of a limit of a function.



To answer the existence question, we must refer to the definition that explains what a limit is. If the conditions specified by the definition are met, then the limit does exist. But, what is meant by a limit?

We consider how values in the range of a function are affected by small changes in values of the domain variable. If we take a number $L$L, we check whether it is possible to find range values as close as we like to $L$L by keeping the corresponding domain values within a small interval surrounding some number $x_0$x0. After tidying up what we mean by 'close to' and 'a small interval' we can then say the function $f(x)$f(x) approaches $L$L as $x$x approaches $x_0$x0.

This is notated $f(x)\rightarrow L$f(x)L as $x\rightarrow x_0$xx0 and $L$L is called the limit as $x$x tends to $x_0$x0. Or, we write $\lim_{x\rightarrow x_0}f(x)=L$limxx0f(x)=L.


  • A limit at a point $x_0$x0 can fail to exist if the function grows larger without bound as $x\rightarrow x_0$xx0. With a slight abuse of notation, we might write $f(x)\rightarrow\infty$f(x) as $x\rightarrow x_0$xx0. But $\infty$ is not a limit because it is not a number.

    This happens, for example, for the function $\tan x$tanx as $x\rightarrow\frac{\pi}{2}$xπ2
  • A limit at a point $x_0$x0 in the domain can also fail to exist if the function values fluctuate rapidly near that point. You could check with a graphing calculator or other software the behaviour of the function defined by $f(x)=\sin\frac{1}{x}$f(x)=sin1x near $x=0$x=0. In this case, if we pick a number $x$x very close to $0$0 it is impossible to guess what the function value will be except to say that it is between $-1$1 and $1$1. The function does not approach a limit at this point of the domain.
  • A limit fails to exist at a point if different values are reached when approaching the point from below or from above. We can, instead, speak of a left-hand limit as $x\rightarrow x_0^-$xx0 or a right-hand limit as $x\rightarrow x_0^+$xx+0.
  • A limit fails to exist at the endpoints of a closed interval. This is because we need to be able to approach a point from both the left and from the right in determining a limit value.



Once we are satisfied that a limit exists at a certain point, we can evaluate it.

If we know that the function is continuous, at a point $x_0$x0, then the limit there is simply $f\left(x_0\right)$f(x0). For example, given the function defined by $f(x)=x^2+1$f(x)=x2+1, which we know to be a continuous function, we can find the limit as $x\rightarrow-2$x2 by evaluating $f(-2)$f(2). That is, the limit $L$L is $(-2)^2+1=5$(2)2+1=5.

There is a circularity trap here because the idea of continuity in functions is defined rigorously in terms of limits. But, for now, we can understand a continuous function to be one that has no sudden jumps in value and no missing values.

The problem of evaluating a limit at a point $x_0$x0 becomes more difficult when $f(x_0)$f(x0) leads to an indeterminate expression like $\frac{0}{0}$00. For example, if we try to find the limit $\lim_{x\rightarrow1}\frac{x^2-3x+1}{x-1}$limx1x23x+1x1 we arrive at $\frac{0}{0}$00 on substituting $x=1$x=1

This happens because the function is not continuous. It has a missing value at $x=1$x=1. However, as long as $x$x is not quite $1$1, we can validly divide the numerator by the denominator and obtain the almost equivalent function $g(x)=x-2$g(x)=x2$x\ne1$x1. This function is continuous everywhere except at $x=1$x=1 where it is undefined. However, the missing value in the function $g$g can be cured by defining $g(1)=-1$g(1)=1.  We can let $x$x approach $1$1 and by evaluating $g(1)$g(1) we see that as $x\rightarrow1$x1, $g(x)\rightarrow-1$g(x)1 and therefore, $f(x)\rightarrow-1$f(x)1

In this case, the limit exists even though the function is undefined at $x=1$x=1


Example 1

The function $g$g is defined by $g(x)=x^2-3$g(x)=x23 over the real numbers.  Show that a limit exists at every point in the domain.

We must convince ourselves that $g(x)$g(x) is close to $g(x_0)$g(x0) whenever $x$x is sufficiently close to $x_0$x0. for each possible $x_0$x0 in the domain. This is always the case for polynomial functions like $g(x)$g(x). We note that none of the four ways listed above for a limit to fail to exist applies in the case of $g(x)$g(x). To be completely rigorous, however, we might go through an argument like the following.

We want to be able to guarantee that $|g(x)-g(x_0)|<\epsilon$|g(x)g(x0)|<ϵ, where $\epsilon$ϵ is a chosen small number, by making $|x-x_0|$|xx0| small enough. The quantity $|x-x_0|$|xx0| is usually given the label $\delta$δ. We require


$\therefore\ \ |x+x_0||x-x_0|<\epsilon$  |x+x0||xx0|<ϵ

$\therefore\ \ |x-x_0|<\frac{\epsilon}{|x+x_0|}$  |xx0|<ϵ|x+x0|

The idea now is that we constrain $x$x so that $|x-x_0|=\delta$|xx0|=δ is small enough and so that a suitable value for $\delta$δ can be given in a way that depends only on $\epsilon$ϵ and $x_0$x0. We look at the term $|x+x_0|$|x+x0|. One way to proceed is to reason that we can restrict $x$x so that it is at most $1$1 unit away from $x_0$x0. That is, $|x-x_0|\le1$|xx0|1. Thus, we have arranged to have $|x+x_0|\le|2x_0+1|$|x+x0||2x0+1|.

To convince yourself that this last step is true you could draw a number-line diagram or argue as follows:

If $|x-x_0|\le1$|xx0|1, then $-1\le x-x_0\le1$1xx01.
$\therefore\ \ -2x_0-1\le x+x_0\le1+2x_0$  2x01x+x01+2x0
$\therefore\ \ x+x_0\le\left|2x_0+1\right|$  x+x0|2x0+1|
$\therefore\ \ \left|x+x_0\right|\le\left|2x_0+1\right|$  |x+x0||2x0+1|

Now, if $|x-x_0|<\frac{\epsilon}{|2x_0+1|}$|xx0|<ϵ|2x0+1|, it will be true that $|x-x_0|<\frac{\epsilon}{|2x_0+1|}<\frac{\epsilon}{|x+x_0|}$|xx0|<ϵ|2x0+1|<ϵ|x+x0| as required. 

We have deduced that for any $x_0$x0 and chosen small value for $\epsilon$ϵ, we can make $|g(x)-g(x_0)|<\epsilon$|g(x)g(x0)|<ϵ by making the distance $|x-x_0|<\delta=\frac{\epsilon}{|2x_0+1|}$|xx0|<δ=ϵ|2x0+1|.

For example, at $x_0=4$x0=4 we would need $\delta=\frac{\epsilon}{9}$δ=ϵ9. If we chose $\epsilon=0.1$ϵ=0.1, then $\delta$δ can be any number less than $\frac{0.1}{9}$0.19. For convenience, say $\delta=0.01$δ=0.01. Now, $g(4-0.01)=12.9201$g(40.01)=12.9201 and $g(4+0.01)=13.0801$g(4+0.01)=13.0801 while $g(4)=13$g(4)=13. We see that $g(x)$g(x) is within $0.1$0.1 of $g(4)$g(4) if $x$x is within $0.01$0.01 of $4$4.

Thus, we can use this formula find a suitable $\delta=|x-x_0|$δ=|xx0| for any $\epsilon$ϵ, however small and this shows that the limit $\lim_{x\rightarrow x_0}$limxx0 exists for every $x_0$x0. It is equal to $g(x_0)$g(x0).


Example 2

Does the rational function $f(x)=\frac{x^2-5}{x-1}$f(x)=x25x1 have a limit at $x=1$x=1?

If we try to evaluate $f(1)$f(1), we obtain $\frac{-4}{0}$40, which is undefined. We could try rewriting $f(x)=\frac{x^2-5}{x-1}=x+1-\frac{4}{x-1}$f(x)=x25x1=x+14x1 but still the fractional part is undefined at $x=1$x=1. This rational function fails to have a limit at $x-1$x1 because it increases without bound as $x\rightarrow1$x1.


Example 3

Does the limit  $\lim_{x\rightarrow1}\frac{x^2+4x-5}{x-1}$limx1x2+4x5x1 exist? If so, what is it?

At $x=1$x=1 the fraction becomes $\frac{0}{0}$00, which is indeterminate. However, we can factorise the numerator and obtain $\frac{(x-1)(x+5)}{x-1}$(x1)(x+5)x1 which is almost everywhere equivalent to $x+5$x+5. The function has a missing value at $x=1$x=1 but we conclude that $\lim_{x\rightarrow1}\frac{x^2+4x-5}{x-1}=6$limx1x2+4x5x1=6.




Worked Examples

Question 1


  1. Does the above limit exist?








  2. What is the value of the limit?

Question 2


$x$x $-5.1$5.1 $-5.01$5.01 $-5.001$5.001 $-5$5 $-4.999$4.999 $-4.99$4.99 $-4.9$4.9
$\frac{x^2+4}{x+5}$x2+4x+5       $\ast$      
  1. Does the above limit exist?









Question 3



Identify discontinuities and limits of functions


Apply differentiation methods in solving problems

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