14.02 Evaluating limits and limit theorems

Evaluation of limits using limit theorems 

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.

1. The limit of a sum or difference is the same as the sum or difference of the limits of each part.  
   $\lim_{x\rightarrow a}f\left(x\right)\pm g\left(x\right)=\lim_{x\rightarrow a}f\left(x\right)\pm\lim_{x\rightarrow a}g\left(x\right)$limx→af(x)±g(x)=limx→af(x)±limx→ag(x)
2. The limit of a constant multiple of a function is equal to a constant multiple of the limit of a function. 
   $\lim_{x\rightarrow a}\left(k\cdot f\left(x\right)\right)=k\cdot\lim_{x\rightarrow a}f\left(x\right)$limx→a(k·f(x))=k·limx→af(x)
3. The limit of a product/quotient is the same as the product/quotient of the limits of each function.
   $\lim_{x\rightarrow a}\frac{f\left(x\right)}{g\left(x\right)}=\frac{\lim_{x\rightarrow a}f\left(x\right)}{\lim_{x\rightarrow a}g\left(x\right)}$limx→af(x)g(x)​=limx→af(x)limx→ag(x)​
4. For the constant function,
   $f\left(x\right)=k,\lim_{x\rightarrow a}f\left(x\right)=k$f(x)=k,limx→af(x)=k

 

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.

 

Worked examples

Question 1

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=log2​64, which is $6$6.

Reflect: Would this method work if we were asked to find the limit as y approaches a negative value?

Question 2

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.

 

 

Existence of a limit

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→x0​f(x)=L.

 

• A limit at a point $x_0$x0​ can fail to exist if the function grows larger without bound as $x\rightarrow x_0$x→x0​. With a slight abuse of notation, we might write $f(x)\rightarrow\infty$f(x)→∞ as $x\rightarrow x_0$x→x0​. But $\infty$∞ is not a limit because it is not a number.
  
  This happens, for example, for the function $\tan x$tanx as $x\rightarrow\frac{\pi}{2}$x→π2​. 
   
• 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 behavour 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 $x\rightarrow x_0^-$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.

 

Evaluation

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. 

 

Exploration

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|.

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​).

 

Practice questions

Question 3

Question 4

Question 5

Question 6