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.

 

Existence

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 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 $x\rightarrow x_0^-$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 $\lim_{x\rightarrow1}\frac{x^2-3x+1}{x-1}$limx→1x2−3x+1x−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. 

 

Example 1

The function $g$g is defined by $g(x)=x^2-3$g(x)=x2−3 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(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 $|x-x_0|$|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​).

 

Example 2

Does the rational function $f(x)=\frac{x^2-5}{x-1}$f(x)=x2−5x−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)=x2−5x−1​=x+1−4x−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.

 

Example 3

Does the limit  $\lim_{x\rightarrow1}\frac{x^2+4x-5}{x-1}$limx→1x2+4x−5x−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 $\lim_{x\rightarrow1}\frac{x^2+4x-5}{x-1}=6$limx→1x2+4x−5x−1​=6.

 

 

 

Worked Examples

Question 1

Question 2

Question 3