NZ Level 8 (NZC) Level 3 (NCEA) [In development]

Evaluating Limits

Lesson

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 $x_0$`x`0. 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 $x_0$`x`0.

This is notated $f(x)\rightarrow L$`f`(`x`)→`L` as $x\rightarrow x_0$`x`→`x`0 and $L$`L` is called the limit as $x$`x` tends to $x_0$`x`0. Or, we write $\lim_{x\rightarrow x_0}f(x)=L$lim`x`→`x`0`f`(`x`)=`L`.

- A limit at a point $x_0$
`x`0 can fail to exist if the function grows larger without bound as $x\rightarrow x_0$`x`→`x`0. With a slight abuse of notation, we might write $f(x)\rightarrow\infty$`f`(`x`)→∞ as $x\rightarrow x_0$`x`→`x`0. But $\infty$∞ is not a limit because it is not a number.

This happens, for example, for the function $\tan x$`t``a``n``x`as $x\rightarrow\frac{\pi}{2}$`x`→π2.

- A limit at a point $x_0$
`x`0 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`)=`s``i``n`1`x` 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 $x\rightarrow x_0^-$
`x`→`x`−0 or a right-hand limit as $x\rightarrow x_0^+$`x`→`x`+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 $x_0$`x`0, then the limit there is simply $f\left(x_0\right)$`f`(`x`0). For example, given the function defined by $f(x)=x^2+1$`f`(`x`)=`x`2+1, which we know to be a continuous function, we can find the limit as $x\rightarrow-2$`x`→−2 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 $x_0$`x`0 becomes more difficult when $f(x_0)$`f`(`x`0) leads to an indeterminate expression like $\frac{0}{0}$00. For example, if we try to find the limit $\lim_{x\rightarrow1}\frac{x^2-3x+1}{x-1}$lim`x`→1`x`2−3`x`+1`x`−1 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`)=`x`−2, $x\ne1$`x`≠1. 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 $x\rightarrow1$`x`→1, $g(x)\rightarrow-1$`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.

The function $g$`g` is defined by $g(x)=x^2-3$`g`(`x`)=`x`2−3 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`(`x`0) whenever $x$`x` is sufficiently close to $x_0$`x`0. for each possible $x_0$`x`0 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`(`x`0)|<`ϵ`, where $\epsilon$`ϵ` is a chosen small number, by making $|x-x_0|$|`x`−`x`0| small enough. The quantity $|x-x_0|$|`x`−`x`0| is usually given the label $\delta$`δ`. We require

$\left|x^2-3-(x_0^2-3)\right|<\epsilon$|`x`2−3−(`x`20−3)|<`ϵ`

$\therefore\ \ |x+x_0||x-x_0|<\epsilon$∴ |`x`+`x`0||`x`−`x`0|<`ϵ`

$\therefore\ \ |x-x_0|<\frac{\epsilon}{|x+x_0|}$∴ |`x`−`x`0|<`ϵ`|`x`+`x`0|

The idea now is that we constrain $x$`x` so that $|x-x_0|=\delta$|`x`−`x`0|=`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 $x_0$`x`0. We look at the term $|x+x_0|$|`x`+`x`0|. One way to proceed is to reason that we can restrict $x$`x` so that it is at most $1$1 unit away from $x_0$`x`0. That is, $|x-x_0|\le1$|`x`−`x`0|≤1. Thus, we have arranged to have $|x+x_0|\le|2x_0+1|$|`x`+`x`0|≤|2`x`0+1|.

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

If $|x-x_0|\le1$|`x`−`x`0|≤1, then $-1\le x-x_0\le1$−1≤`x`−`x`0≤1.

$\therefore\ \ -2x_0-1\le x+x_0\le1+2x_0$∴ −2`x`0−1≤`x`+`x`0≤1+2`x`0

$\therefore\ \ x+x_0\le\left|2x_0+1\right|$∴ `x`+`x`0≤|2`x`0+1|

$\therefore\ \ \left|x+x_0\right|\le\left|2x_0+1\right|$∴ |`x`+`x`0|≤|2`x`0+1|

Now, if $|x-x_0|<\frac{\epsilon}{|2x_0+1|}$|`x`−`x`0|<`ϵ`|2`x`0+1|, it will be true that $|x-x_0|<\frac{\epsilon}{|2x_0+1|}<\frac{\epsilon}{|x+x_0|}$|`x`−`x`0|<`ϵ`|2`x`0+1|<`ϵ`|`x`+`x`0| as required.

We have deduced that for any $x_0$`x`0 and chosen small value for $\epsilon$`ϵ`, we can make $|g(x)-g(x_0)|<\epsilon$|`g`(`x`)−`g`(`x`0)|<`ϵ` by making the distance $|x-x_0|<\delta=\frac{\epsilon}{|2x_0+1|}$|`x`−`x`0|<`δ`=`ϵ`|2`x`0+1|.

For example, at $x_0=4$`x`0=4 we would need $\delta=\frac{\epsilon}{9}$`δ`=`ϵ`9. 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`(4−0.01)=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`=|`x`−`x`0| for any $\epsilon$`ϵ`, however small and this shows that the limit $\lim_{x\rightarrow x_0}$lim`x`→`x`0 exists for every $x_0$`x`0. It is equal to $g(x_0)$`g`(`x`0).

Does the rational function $f(x)=\frac{x^2-5}{x-1}$`f`(`x`)=`x`2−5`x`−1 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`)=`x`2−5`x`−1=`x`+1−4`x`−1 but still the fractional part is undefined at $x=1$`x`=1. This rational function fails to have a limit at $x-1$`x`−1 because it increases without bound as $x\rightarrow1$`x`→1.

Does the limit $\lim_{x\rightarrow1}\frac{x^2+4x-5}{x-1}$lim`x`→1`x`2+4`x`−5`x`−1 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 $\frac{(x-1)(x+5)}{x-1}$(`x`−1)(`x`+5)`x`−1 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 $\lim_{x\rightarrow1}\frac{x^2+4x-5}{x-1}=6$lim`x`→1`x`2+4`x`−5`x`−1=6.

$\lim_{x\to-2}\sqrt{2x+5}$lim`x`→−2√2`x`+5

Does the above limit exist?

Yes

ANo

BYes

ANo

BWhat is the value of the limit?

$\lim_{x\to-5}\left(\frac{x^2+4}{x+5}\right)$lim`x`→−5(`x`2+4`x`+5)

$x$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 |
$\ast$∗ |

Does the above limit exist?

Yes

ANo

BYes

ANo

B

Identify discontinuities and limits of functions

Apply differentiation methods in solving problems