There are a number of limit laws which we tend to use in limit questions without writing them into our calculations in a formal manner.
For well-behaved functions that don't have jumps or breaks near where we're finding the limit, finding a limit at a point in the domain is usually a matter of evaluating the function at that point.
So for a function $f\left(x\right)$f(x) that has no jumps around $x=a$x=a:
$\lim_{x\to a}f\left(x\right)=f\left(a\right)$limx→af(x)=f(a)
Other functions may have points where they are undefined, which can mean a limit does not exist at those points, or it can happen that a limit does exist at such points and we need to use the limit theorems given above combined with knowledge of previously discovered limits to evaluate it.
Evaluate $\lim_{x\rightarrow25}\log_2\left(59+\sqrt{x}\right)$limx→25log2(59+√x).
Think: The function given by $\log_2\left(59+\sqrt{x}\right)$log2(59+√x) is defined everywhere on the domain $x>0$x>0 with no jumps.
Do: We substitute $25$25 for $x$x and simplify to obtain $\log_2(59+\sqrt{25}=\log_264$log2(59+√25=log264, which is $6$6.
Reflect: Would this method work if we were asked to find the limit as y approaches a negative value?
Evaluate $\lim_{t\rightarrow0}\frac{\sin^2t+t^3}{t^2}$limt→0sin2t+t3t2.
Think: Substitution of $t=0$t=0 gives the indeterminate form $\frac{0}{0}$00. So, we use the limit theorems.
Do: Using theorem $1$1, we write the statement as $\lim_{t\rightarrow0}\frac{\sin^2t}{t^2}+\lim_{t\rightarrow0}\frac{t^3}{t^2}$limt→0sin2tt2+limt→0t3t2.
We split this further with the help of the third theorem.
$\lim_{t\rightarrow0}\frac{\sin t}{t}.\lim_{t\rightarrow0}\frac{\sin t}{t}+\lim_{t\rightarrow0}\frac{t^3}{t^2}$limt→0sintt.limt→0sintt+limt→0t3t2.
Now, we evaluate the separate limit statements. For $\lim_{t\rightarrow0}\frac{t^3}{t^2}$limt→0t3t2 we note that when $t$t is very near but not equal to $0$0, the statement is equivalent to $\lim_{t\rightarrow0}\ t$limt→0 t which is $0$0. So, we have
$\lim_{t\rightarrow0}\frac{\sin^2t+t^3}{t^2}=1\times1+0=1$limt→0sin2t+t3t2=1×1+0=1.
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 $x\rightarrow x_0$x→x0 and $L$L is called the limit as $x$x tends to $x_0$x0. Or, we write $\lim_{x\rightarrow x_0}f(x)=L$limx→x0f(x)=L.
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 $x\rightarrow-2$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 $x_0$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 $\lim_{x\rightarrow1}\frac{x^2-3x+2}{x-1}$limx→1x2−3x+2x−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)=x2−3 over the real numbers. Let's use the definition of a limit to show that a limit exists at every point in the domain.
Think: 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.
Do: 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 $|x-x_0|$|x−x0| small enough. The quantity $|x-x_0|$|x−x0| is usually given the label $\delta$δ. We require
$\left|x^2-3-(x_0^2-3)\right|<\epsilon$|x2−3−(x20−3)|<ϵ
$\therefore\ \ |x+x_0||x-x_0|<\epsilon$∴ |x+x0||x−x0|<ϵ
$\therefore\ \ |x-x_0|<\frac{\epsilon}{|x+x_0|}$∴ |x−x0|<ϵ|x+x0|
The idea now is that we constrain $x$x so that $|x-x_0|=\delta$|x−x0|=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, $|x-x_0|\le1$|x−x0|≤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 $|x-x_0|\le1$|x−x0|≤1, then $-1\le x-x_0\le1$−1≤x−x0≤1.
$\therefore\ \ -2x_0-1\le x+x_0\le1+2x_0$∴ −2x0−1≤x+x0≤1+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 $|x-x_0|<\frac{\epsilon}{|2x_0+1|}$|x−x0|<ϵ|2x0+1|, it will be true that $|x-x_0|<\frac{\epsilon}{|2x_0+1|}<\frac{\epsilon}{|x+x_0|}$|x−x0|<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 $|x-x_0|<\delta=\frac{\epsilon}{|2x_0+1|}$|x−x0|<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(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−x0| for any $\epsilon$ϵ, however small and this shows that the limit $\lim_{x\rightarrow x_0}$limx→x0 exists for every $x_0$x0. It is equal to $g(x_0)$g(x0).
For many functions, we can evaluate the limit $\lim_{x\to a}f(x)$limx→af(x) by looking at the graph, a table of values, or substituting in $f(a)$f(a).
For some functions, such as ones where $f(a)$f(a) is undefined, we may need to look at both the right and left side limits to ensure the limit exists, so
If $\lim_{x\to3}f\left(x\right)=7$limx→3f(x)=7, find $\lim_{x\to3}\left(1+f\left(x\right)\right)^2$limx→3(1+f(x))2.
$\lim_{x\to-2}\sqrt{2x+5}$limx→−2√2x+5
Does the above limit exist?
Yes
No
What is the value of the limit?
$\lim_{x\to-5}\left(\frac{x^2+4}{x+5}\right)$limx→−5(x2+4x+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
No
The graph of $y=\frac{x+3}{\left(x-5\right)^2}$y=x+3(x−5)2 is pictured below.
Does the limit $\lim_{x\to5}\left(\frac{x+3}{\left(x-5\right)^2}\right)$limx→5(x+3(x−5)2) exist?
Yes
No