Find Limits from a Graph or Table of Values
Although we will need a rigorous method of determining that a limit exists at a given point in the domain of a function, and what its value is, we can often resolve this issue confidently by looking at a graph or by constructing a table of values for points close to the point in question.
If a function reaches arbitrarily large positive or negative values as some value of the variable is approached, then no limit exists at that point. The function is said to have a vertical asymptote at that point. The function is undefined there and is said to be discontinuous.
If in crossing some point in the domain, the function values jump without passing through intermediate values, we say the function has a jump discontinuity at that point. The limit as the point is approached from below will be different from the limit as the point is approached from above.
We may speak of a left-hand and a right-hand limit in this case but a strict limit does not exist at that point. (The notations $\lim_{x\rightarrow a^-}$limx→a− and $\lim_{x\rightarrow a^+}$limx→a+ are used for these left- and right-hand limits.)
A function can fail to be continuous at some point but still have a limit there. The following graph illustrates such a case. The function value need not be the same as the value of the limit at a given point.
Graphs and tables may be sufficient to allow the conclusion that a limit does not exist at a given point but to be certain that a limit does exist more care is needed. The thinking is essentially the following:
Suppose a number $L$L is thought to be the limiting function value as a certain point $a$a in the domain is approached. To be certain that the limit exists, we require that the function attains values close to $L$L when the variable is sufficiently close to $a$a. This property has to hold however small we choose the distance of a function value to be from $L$L.
If the function has values arbitrarily close to $L$L for domain values sufficiently close to $a$a, then we can conclude that the limit as $a$a is approached is $L$L.
Roughly speaking, if the graph of a function sufficiently close to but surrounding a point looks like a straight line graph, then the function has a limit at that point and it is equal to the function value if it exists.
Example 1
Consider the quadratic function defined on the interval $(-3,3)$(−3,3) given by $y=\frac{x^2}{2}-x+\frac{1}{2}$y=x22−x+12.
At $x=2$x=2, the function value is $\frac{1}{2}$12. It is evident from the graph that if the value $2$2 is approached from below or from above, the corresponding function values approach $\frac{1}{2}$12. We conclude that $\lim_{x\rightarrow2}=\frac{1}{2}$limx→2=12.
Similar limit statements can be made about every point inside the domain interval. In each case, the limit is equal to the function value.
Example 2
This is the graph of the function $f(x)=\frac{\sin2x}{x}$f(x)=sin2xx. This function exists for all real numbers $x$x except $x=0$x=0. However, the graph suggests that there is a limit as $x\rightarrow0$x→0 and that this limit is $2$2.
To investigate this, a table of values could be constructed for points surrounding $x=0$x=0. [The values of the sine function assume that the argument is expressed in radian measure.]
| $x$x | $-0.1$−0.1 | $-0.01$−0.01 | $-0.001$−0.001 | $0.001$0.001 | $0.01$0.01 |
| $\frac{\sin2x}{x}$sin2xx | $1.98669$1.98669 | $1.999866$1.999866 | $1.9999986$1.9999986 | $1.9999986$1.9999986 | $1.999866$1.999866 |
Example 3
Consider the graph of a function, $f\left(x\right)$f(x), shown below.
Does $\lim_{x\to0}f\left(x\right)$limx→0f(x) exist?
Yes, the limit exists.
ANo, the limit does not exist.
B
Example 4
$\lim_{x\to2}\left|3x-4\right|$limx→2|3x−4|
Does the above limit exist?
Yes
ANo
BWhat is the value of the limit?
Example 5
Consider the function $f\left(x\right)=\frac{2-x}{x^2+2}$f(x)=2−xx2+2.
Complete the table to find the values of $f\left(x\right)$f(x) as $x$x gets closer and closer to $0$0 from the left, and closer and closer to $0$0 from the right. Give your answers correct to 4 decimal places.
$x$x $-0.1$−0.1 $-0.01$−0.01 $-0.001$−0.001 $0.001$0.001 $0.01$0.01 $0.1$0.1 $f\left(x\right)$f(x) $\editable{}$ $\editable{}$ $\editable{}$ $\editable{}$ $\editable{}$ $\editable{}$ Hence, find the value of $\lim_{x\to0}\left(\frac{2-x}{x^2+2}\right)$limx→0(2−xx2+2).