Introduction to limits
Definition and introductive examples
The concept of limit is fundamental to the study of calculus. The concept of a limit is to understand the behavior of functions at infinity or near infinitely.
For example, let us consider the following function's behavior as x approaches \infty :
f\left(x\right)=\dfrac{4x+1}{x}
| x | \dfrac{4x+1}{x} |
| .1 | 14 |
| 1 | 5 |
| 10 | 4.1 |
| 100 | 4.01 |
| 1000 | 4.001 |
| 10 000 | 4.0001 |
| 1,000,000 | 4.00001 |
| 10,000,000 | 4.000001 |
| 100,000,000 | 4.0000001 |
Observe as the values of x become larger and larger, the values of f\left(x\right) approach 4.
Definition
Limit of a function
Let f\left(x\right) be a function and L a real number. If the values of f\left(x\right) are near L whenever x approaches a, but are not equal to a, we say that the limit as x approaches a is L. We write:
\lim\limits_{x\to a}f\left(x\right)=L
Consider the following function:
f\left(x\right)=2x+1
As x approaches a=3, the values of f\left(x\right) approach 7 :
| x | 2x+1 |
| 1 | 3 |
| 2 | 5 |
| 2.5 | 6 |
| 2.9 | 6.8 |
| 3 | 7 |
| 3.1 | 7.2 |
| 3.5 | 8 |
| 4 | 9 |
Observe for x -values near 3 that the values of 2x+1 are near 7. In particular,
\lim\limits_{x\to 3}\, 2x+1=7
If there is no number L so that the limit as x approaches a is L, then we say the limit of f\left(x\right) does not exist at a.
Consider the following function:
f\left(x\right)=\dfrac{1}{x}
Observe for positive x -values near 0 that the function \dfrac{1}{x} takes on large positive values. However, for negative x -values near 0, the function \dfrac{1}{x} takes on large negative values. Therefore, the limit \lim\limits_{x\to 0}\dfrac{1}{x} does not exist. This can also be seen graphically. Consider the graph of \dfrac{1}{x} :
Observe as x approaches 0 from the right that the graph of f\left(x\right) approaches \infty and as x approaches 0 from the left that the graph of f\left(x\right) approaches -\infty. Therefore the limit does not exist.
Limit at infinity
Let f\left(x\right) be a function and L a real number. If the values of f\left(x\right) are arbitrarily near L whenever x is arbitrarily large and positive, we write:
\lim\limits_{x\to \infty }f\left(x\right)=L
Consider the following function:
f\left(x\right)=\dfrac{1}{x}
We have the following:
\lim\limits_{x\to \infty}\dfrac{1}{x}=0
The above limit at infinity is zero because, for large values of x, the denominator of the rational function is large while the numerator of the rational function is the relatively small number 1.
Let f\left(x\right) be a function and L a real number. If the values of f\left(x\right) are arbitrarily near L whenever x is arbitrarily large and negative, then we write:
\lim\limits_{x\to -\infty }f\left(x\right)=L
Lefthand and righthand limits
Lefthand limit
Let f\left(x\right) be a function and L a real number. If the values of f\left(x\right) are near L whenever x is slightly less than but not equal to a, we say that the lefthand limit as x approaches a is L. We write:
\lim\limits_{x\to a^-}f\left(x\right)=L
Consider the greatest integer function:
f\left(x\right)=\lfloor x \rfloor
Recall that if a is a real number, then \lfloor a\rfloor is the greatest integer which is less than or equal to a.
In particular, for x -values satisfying 0\leq x\lt 1 we have:
\lfloor x\rfloor=0
Therefore:
\lim\limits_{x\to 1^-}f\left(x\right)=0
Righthand limit
Let f\left(x\right) be a function and L a real number. If the values of f\left(x\right) are near L whenever x is slightly greater than but not equal to a, we say that the righthand limit as x approaches a is L. We write:
\lim\limits_{x\to a^+}f\left(x\right)=L
Consider the greatest integer function:
f\left(x\right)=\lfloor x \rfloor
Work from the previous examples proves the following:
\lim\limits_{x\to 1^+}f\left(x\right)=1
The information about the limits of the floor function can also be seen graphically:
For x -value slightly less than 1, the graph agrees with y=0. For x -values slightly greater than 1, the graph agrees with y=1.
The limit of a function at a point exists if and only if the lefthand and righthand limits exist and agree at that point.
Let f\left(x\right) be a function and L a real number. Then \lim\limits_{x\to a}f\left(x\right)=L if and only if:
\lim\limits_{x\to a^-}f\left(x\right)=\lim\limits_{x\to a^+}f\left(x\right)=L
Limits and graphs
Graphic interpretation for a limit at infinity
Suppose f\left(x\right) is a function, L is a real number, and we have the following limit at infinity:
\lim\limits_{x\to \infty}f\left(x\right)=L
The graph of f\left(x\right) will approach the line y=L as x approaches infinity. Equivalently, the graph of f\left(x\right) will have a horizontal asymptote of y=L.
Consider the arctangent function:
f\left(x\right)=\arctan\left(x\right)
By examining the graph of f\left(x\right), we have the following limits at infinity:
- \lim\limits_{x\to \infty}\arctan\left(x\right)=\dfrac{\pi}{2}
- \lim\limits_{x\to -\infty}\arctan\left(x\right)=\dfrac{-\pi}{2}
The graph of the arctangent function f\left(x\right)=\arctan\left(x\right) has horizontal asymptotes of y=\dfrac{\pi}{2} and y=-\dfrac{\pi}{2}.
Suppose f\left(x\right) is a function and we have the following limit at infinity:
\lim\limits_{x\to \pm\infty}f\left(x\right)=\pm \infty
Then the graph will not have a horizontal asymptote at \infty.
Consider the quadratic function:
f\left(x\right)=x^2
By examining the graph of f\left(x\right), we have the following limits at infinity:
- \lim\limits_{x\to \infty}x^2=\infty
- \lim\limits_{x\to -\infty}x^2=\infty
Graphic interpretation for a limit at a point
Suppose f\left(x\right) is a function and L is a real number. Suppose further that f is not defined at point a and that we have the following limit:
\lim\limits_{x\to a}f\left(x\right)=\pm \infty
The graph will have a vertical asymptote at x=a.
Consider the graph of the function f\left(x\right)=\dfrac{1}{x-1} :
Observe that the graph of \dfrac{1}{x-1} has a vertical asymptote at x=1 and that the pieces of the graph do not meet to the left and to the the right of the vertical asymptote.
Limits of usual functions
Limit at a point
Limits of usual functions such as:
- polynomial functions
- trigonometric functions
- exponential functions
- logarithmic functions
at points in their domain can be found by simply evaluating the function at the point.
Consider the following function:
f\left(x\right)=x^2-3x+4
Then:
\lim\limits_{x\to 3}f\left(x\right)=f\left(3\right)=3^2-3\left(3\right)+4=4
Limit at the infinity
Limits at infinity of a polynomial function
Let f\left(x\right)=x^n+a_{n-1}x^{n-1}\cdots a_1x+a_0 be a polynomial function. Then:
\lim\limits_{x\to \infty}f\left(x\right)=\infty
And:
\lim\limits_{x\to -\infty}f\left(x\right)=\begin{cases} \infty &\mbox{if }n\mbox{ is even} \cr \cr -\infty & \mbox{if }n \mbox{ is odd} \end{cases}
Consider the following polynomial function:
f\left(x\right)=x^3+1
Then:
\lim\limits_{x\to \infty} x^3+1=\infty\\\mbox{and}\\\lim\limits_{x\to-\infty}x^3+1=-\infty
Limit at infinity of rational functions
Consider two polynomial functions:
- g\left(x\right)=a_nx^n+\cdots + a_1x+a_0
- h\left(x\right)=b_mx^m+\cdots +b_1x+b_0
Let f\left(x\right) be the following rational function:
f\left(x\right)=\dfrac{g\left(x\right)}{h\left(x\right)}
Then we have the following limits at infinity:
- \lim\limits_{x\to \infty}f\left(x\right)=0 if n \lt m
- \lim\limits_{x\to \infty}f\left(x\right)=\infty if n \gt m
- \lim\limits_{x\to \infty}f\left(x\right)=\dfrac{a_n}{b_m} if n=m
Consider the following rational function:
f\left(x\right)=\dfrac{2x^7-3x^2+110}{-15x^7+12x^3}
The degree of numerator agrees with the degree of the denominator and we therefore have the following limit at infinity:
\lim\limits_{x\to \infty}f\left(x\right)=\dfrac{2}{-15}
Limit at infinity of exponential functions
Let a be a positive real number not equal to 1 and consider the following exponential function:
f\left(x\right)=a^x.
If a \gt 1, then:
\lim\limits_{x\to \infty}a^x=\infty\mbox{ and }\lim\limits_{x\to-\infty}a^x=0
If 0\lt a \lt 1, then:
\lim\limits_{x\to \infty}a^x=0\mbox{ and }\lim\limits_{x\to-\infty}a^x=\infty
Consider the following exponential function:
f\left(x\right)=e^x
The number e is greater than 1. Therefore:
\lim\limits_{x\to \infty}f\left(x\right)=\infty
Limit at infinity of logarithmic functions
Let a be a positive real number not equal to 1 and consider the following logarithmic function:
f\left(x\right)=\log_a\left(x\right)
If a \gt 1, then:
\lim\limits_{x\to \infty}\log_a\left(x\right)=\infty
If 0\lt a \lt 1, then:
\lim\limits_{x\to \infty}\log_a\left(x\right)=-\infty
Consider the natural logarithmic function:
f\left(x\right)=\ln\left(x\right)=\log_e\left(x\right)
The number e is greater than 1. Therefore:
\lim\limits_{x\to \infty}f\left(x\right)=\infty
The periodic nature of trigonometric functions cause most trigonometric functions not to have a limit at \infty.
Consider the graph of trigonometric function f\left(x\right)=\sin\left(x\right) :
As x tends to \infty, the graph of \sin\left(x\right) oscillates between 1 and -1. Hence the function f\left(x\right)=\sin\left(x\right) does not have a limit as x approaches \infty.
Properties of limits
Addition, subtraction, and constant laws
Addition law
Addition law
Suppose f\left(x\right) and g\left(x\right) are functions and that:
- \lim\limits_{x\to a}f\left(x\right)=L_1
- \lim\limits_{x\to a}g\left(x\right)=L_2
Then we have the following limit:
\lim\limits_{x\to a}\left(f\left(x\right)+g\left(x\right)\right)=L_1+L_2
Consider the following two functions:
- f\left(x\right)=\sin\left(x\right)
- g\left(x\right)=\tan\left(x\right)
Then we have the following limits:
- \lim\limits_{x\to \frac{\pi}{4}}\sin\left(x\right)=\dfrac{\sqrt{2}}{2}
- \lim\limits_{x\to \frac{\pi}{4}}\tan\left(x\right)=1
Therefore:
\lim\limits_{x\to \frac{\pi}{4}}\left(\sin\left(x\right)+\tan\left(x\right)\right)=\dfrac{\sqrt{2}}{2}+1
Subtraction law
Subtraction law
Suppose f\left(x\right) and g\left(x\right) are functions and that:
- \lim\limits_{x\to a}f\left(x\right)=L_1
- \lim\limits_{x\to a}g\left(x\right)=L_2
Then we have the following limit:
\lim\limits_{x\to a}\left(f\left(x\right)-g\left(x\right)\right)=L_1-L_2
Consider the following two functions:
- f\left(x\right)=\sin\left(x\right)
- g\left(x\right)=\tan\left(x\right)
Then we have the following limits:
- \lim\limits_{x\to \frac{\pi}{4}}\sin\left(x\right)=\dfrac{\sqrt{2}}{2}
- \lim\limits_{x\to \frac{\pi}{4}}\tan\left(x\right)=1
Therefore:
\lim\limits_{x\to \frac{\pi}{4}}\left(\sin\left(x\right)-\tan\left(x\right)\right)=\dfrac{\sqrt{2}}{2}-1
Constant law
Constant law
Suppose f\left(x\right) is a function and that:
\lim\limits_{x\to a}f\left(x\right)=L_1
If C is a real number, then we have the following limit:
\lim\limits_{x\to a}Cf\left(x\right)=CL_1
Consider the following function:
f\left(x\right)=\arctan\left(x\right)
Then we have:
\lim\limits_{x\to \infty}\arctan\left(x\right)=\dfrac{\pi}{4}
Therefore:
\lim\limits_{x\to \infty}4\arctan\left(x\right)=4\cdot \dfrac{\pi}{4}=\pi
Multiplication and division laws
Multiplication law
Multiplication law
Suppose f\left(x\right) and g\left(x\right) are functions and:
- \lim\limits_{x\to a}f\left(x\right)=L_1
- \lim\limits_{x\to a}g\left(x\right)=L_2
Then we have the following limit:
\lim\limits_{x\to a}\left(f\left(x\right)g\left(x\right)\right)=L_1L_2
Consider the following functions:
- f\left(x\right)=x^2-1
- g\left(x\right)=|x|
Then we have the following:
- \lim\limits_{x\to 0}\left(x^2-1\right)=-1
- \lim\limits_{x\to 0}|x|= 0
It follows that:
\lim\limits_{x\to 0}\left(x^2-1\right)|x|=\left(-1\right)\cdot 0=0
Division law
Division law
Suppose f\left(x\right) and g\left(x\right) are functions and that:
- \lim\limits_{x\to a}f\left(x\right)=L_1
- \lim\limits_{x\to a}g\left(x\right)=L_2
Suppose further that L_2\not = 0. Then we have the following limit:
\lim\limits_{x\to a}\dfrac{f\left(x\right)}{g\left(x\right)}=\dfrac{L_1}{L_2}
Consider the following function:
- f\left(x\right)=\tan\left(x\right)=\dfrac{\sin\left(x\right)}{\cos\left(x\right)}
Then we have the following:
- \lim\limits_{x\to \frac{\pi}{4}}\sin\left(x\right)=\dfrac{\sqrt{2}}{2}
- \lim\limits_{x\to \frac{\pi}{4}}\cos\left(x\right)=\dfrac{\sqrt{2}}{2}
Therefore:
\lim\limits_{x\to \frac{\pi}{4}}\tan\left(x\right)=\dfrac{\dfrac{\sqrt{2}}{2}}{\dfrac{\sqrt{2}}{2}}\\=1
Power, root, and composition laws
Power law
Power law
Suppose f\left(x\right) is a function and that:
\lim\limits_{x\to a}f\left(x\right)=L
Then if n is a positive integer:
\lim\limits_{x\to a}\left(f\left(x\right)\right)^n=L^n
Consider the following function:
f\left(x\right)=\sin\left(x\right)
Then we have the following:
\lim\limits_{x\to \frac{\pi}{3}}\sin\left(x\right)=\dfrac{\sqrt{3}}{2}
It follows:
\lim\limits_{x\to \frac{\pi}{4}}\sin^2\left(x\right)=\left(\dfrac{\sqrt{3}}{2}\right)^2\\=\dfrac{3}{4}
Root law
Root law
Suppose f\left(x\right) is a function and that:
\lim\limits_{x\to a}f\left(x\right)=L
If n is an odd number, then:
\lim\limits_{x\to a}\left(f\left(x\right)\right)^{\frac{1}{n}}=L^{\frac{1}{n}}
If n is an even number, then:
- \lim\limits_{x\to a}\left(f\left(x\right)\right)^{\frac{1}{n}}=L^{\frac{1}{n}} if L \gt 0
- \lim\limits_{x\to a}\left(f\left(x\right)\right)^{\frac{1}{n}} does not exit if L\leq 0
Consider the following piecewise function:
f\left(x\right)=\begin{cases} x^2-1 & x\not= 2 \cr \cr 17 & x=2 \end{cases}
Then we have the following:
\lim\limits_{x\to 2}f\left(x\right)=2^2-1=3
It follows that:
\lim\limits_{x\to 2}\left(f\left(x\right)\right)^{\frac{1}{4}}=3^{\frac{1}{4}}
Composition law
Composition law
Suppose f\left(x\right) and g\left(x\right) are functions and that:
- \lim\limits_{x\to a}f\left(x\right)=L_1
- \lim\limits_{x\to L_1}g\left(x\right)=L_2
Then we have the following limit:
\lim\limits_{x\to a}g\left(f\left(x\right)\right)=L_2
Consider the following functions:
- f\left(x\right)=x-7
- g\left(x\right)=x^2+9
Then we have the following:
- \lim\limits_{x\to 8}f\left(x\right)=1
- \lim\limits_{x\to 1}g\left(x\right)=9
It follows that:
\lim\limits_{x\to 8}g\left(f\left(x\right)\right)=9
Find a zero-denominator limit using factorization
Suppose f\left(x\right)=\dfrac{g\left(x\right)}{h\left(x\right)} is a rational function and that a is a real number such that h\left(a\right)=0.
We can find the limit by using factorization and simplifying the rational function.
Consider the following rational function:
f\left(x\right)=\dfrac{x^2-1}{x^2-3x+2}
Observe that the denominator of the rational function is 0 when evaluated at x=1. However, the rational function can be reduced:
f\left(x\right)=\dfrac{x^2-1}{x^2-3x+2}\\=\dfrac{\left(x-1\right)\left(x+1\right)}{\left(x-1\right)\left(x-2\right)}\\=\dfrac{x+1}{x-2}
Therefore we have the following limit:
\lim\limits_{x\to 1}f\left(x\right)=\lim\limits_{x\to 1}\dfrac{x+1}{x-2}=\dfrac{2}{-1}\\=-2
L'Hospital's rule
L'Hospital's
Let g\left(x\right) and h\left(x\right) be differentiable functions and suppose that for some real number a :
- \lim\limits_{x\to a}g\left(x\right)=\lim\limits_{x\to a}h\left(x\right)=0 or
- \lim\limits_{x\to a}g\left(x\right)=\lim\limits_{x\to a}h\left(x\right)=\infty
Then we have the following:
\lim\limits_{x\to a}\dfrac{g\left(x\right)}{h\left(x\right)}=\lim\limits_{x\to a}\dfrac{g'\left(x\right)}{h'\left(x\right)}
Reconsider the following rational function:
f\left(x\right)=\dfrac{x^2-1}{x^2-3x+2}
Then we have the following:
- \lim\limits_{x\to 1}\left(x^2-1\right)=0
- \lim\limits_{x\to1}\left(x^2-3x+2\right)=0
By L'Hospital's rule we therefore have:
\lim\limits_{x\to 1}\dfrac{x^2-1}{x^2-3x+2}=\lim\limits_{x\to 1}\dfrac{\left(x^2-1\right)'}{\left(x^2-3x+2\right)'}\\=\lim\limits_{x\to 1}\dfrac{2x}{2x-3}\\=\dfrac{2}{-1}\\=-2
Consider the following function:
f\left(x\right)=\dfrac{\sin\left(x\right)}{x}
Then we have the following:
- \lim\limits_{x\to 0} \sin\left(x\right)=0\\
- \lim\limits_{x\to 0}x=0
By L'Hospital's rule:
\lim\limits_{x\to 0}\dfrac{\sin\left(x\right)}{x}=\lim\limits_{x\to 0}\dfrac{\cos\left(x\right)}{1}\\=\cos\left(0\right)\\=1
L'Hospital's rule can even be used multiple times in a single example to compute a limit.
Consider the following rational function:
f\left(x\right)=\dfrac{x^2-1}{x^3-x^2-x+1}
Then we have the following:
- \lim\limits_{x\to 1} \left(x^2-2x+1\right)=0
- \lim\limits_{x\to 1}\left(x^3-x^2-x+1\right)=0
Therefore by L'Hospital's rule:
\lim\limits_{x\to 1}\dfrac{x^2-2x+1}{x^3-x^2-x+1}=\lim\limits_{x\to 1}\dfrac{2x-2}{3x^2-2x-1}
Now observe the following:
- \lim\limits_{x\to 1}\left(2x-2\right)=0
- \lim\limits_{x\to 1}\left(3x^2-2x-1\right)=0
Applying L'Hospital's rule a second time:
\lim\limits_{x\to 1}\dfrac{x^2-2x+1}{x^3-x^2-x+1}=\lim\limits_{x\to 1}\dfrac{2x-2}{3x^2-2x-1}\\\lim\limits_{x\to 1}\dfrac{2}{6x-2}\\=\dfrac{2}{4}\\=\dfrac{1}{2}