Limits of Functions

Thelimitof a function at a point \(a\) in its domain (if it exists) is the value that the function approaches as its argument approaches \(a.\) The concept of a limit is the fundamental concept of calculus and analysis. It is used to define thederivativeand thedefinite integral, and it can also be used to analyze the local behavior of functions near points of interest.

Informally, a function is said to have a limit \( L \) at \( a \) if it is possible to make the function arbitrarily close to \( L \) by choosing values closer and closer to \( a \). Note that the actual value at \( a \) is irrelevant to the value of the limit.

The notation is as follows:

\[ \lim_{x \to a} f(x) = L, \]

which is read as "the limit of \(f(x) \) as \(x\) approaches \(a\) is \(L.\)"

The limit of \( f(x) \) at \(x_0\) is the \(y\)-coordinate of the red point, not \(f(x_0).\)^[1]

Main Article:Epsilon-Delta Definition of a Limit

The precise definition of the limit is discussed in the wikiEpsilon-Delta Definition of a Limit.

Formal Definition of a Function Limit:

The limit of \(f(x)\) as \(x\) approaches \(x_0\) is \(L\), i.e.

\[\lim _{ x \to x_{0} }{f(x) } = L\]

if, for every \(\epsilon > 0 \), there exists \(\delta >0 \) such that, for all \(x\),

\[ 0 < \left| x - x_{0} \right |<\delta \textrm{ } \implies \textrm{ } \left |f(x) - L \right| < \epsilon. \]

In practice, this definition is only used in relatively unusual situations. For many applications, it is easier to use the definition to prove some basic properties of limits and to use those properties to answer straightforward questions involving limits.

The most important properties of limits are thealgebraicproperties, which say essentially that limits respect algebraic operations:

Suppose that \( \lim\limits_{x\to a} f(x) = M\) and \(\lim\limits_{x\to a} g(x) = N.\) Then

\[ \begin{align} \lim\limits_{x\to a} \big(f(x)+g(x)\big) &= M+N \\\\ \lim\limits_{x\to a} \big(f(x)-g(x)\big) &= M-N \\\\ \lim\limits_{x\to a} \big(f(x)g(x)\big) &= MN \\\\ \lim\limits_{x\to a} \left(\frac{f(x)}{g(x)}\right) &= \frac MN \ \ \text{ (if } N\ne 0) \\\\ \lim\limits_{x\to a} f(x)^k &= M^k \ \ \text{ (if } M,k > 0). \end{align} \]

These can all be proved via application of the epsilon-delta definition. Note that the results are only true if the limits of the individual functions exist: if \( \lim\limits_{x\to a} f(x) \) and \( \lim\limits_{x\to a} g(x)\) do not exist, the limit of their sum (or difference, product, or quotient) might nevertheless exist.

Coupled with the basic limits \( \lim_{x\to a} c = c,\) where \( c\) is a constant, and \( \lim_{x\to a} x = a,\) the properties can be used to deduce limits involvingrational functions:

Let \( f(x) \) and \(g(x)\) be polynomials, and suppose \(g(a) \ne 0.\) Then

\[ \lim_{x\to a} \frac{f(x)}{g(x)} = \frac{f(a)}{g(a)}. \]

This is an example ofcontinuity, or what is sometimes called limits bysubstitution.

Note that \(g(a)=0\) is a more difficult case; see theIndeterminate Formswiki for further discussion.

Let \(m\) and \(n\) be positive integers. Find

\[ \lim_{x\to 1} \frac{x^m-1}{x^n-1}. \]

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Immediately substituting \(x=1\) does not work, since the denominator evaluates to \(0.\) First, divide top and bottom by \(x-1\) to get

\[\frac{x^{m-1}+x^{m-2}+\cdots+1}{x^{n-1}+x^{n-2}+\cdots+1}.\]

Plugging in \(x=1\) to the denominator does not give \(0,\) so the limit is this fraction evaluated at \(x=1,\) which is

\[\frac{1^{m-1}+1^{m-2}+\cdots+1}{1^{n-1}+1^{n-2}+\cdots+1} = \frac{m}{n}.\ _\square\]

It is important to notice that the manipulations in the above example are justified by the fact that \( \lim\limits_{x\to a} f(x)\) is independent of the value of \(f(x) \) at \(x=a,\) or whether that value exists. This justifies, for instance, dividing the top and bottom of the fraction \(\frac{x^m-1}{x^n-1}\) by \(x-1,\) since this is nonzero for \(x\ne 1.\)

A片面的limit only considers values of a function that approaches a value from either above or below.

The right-side limit of a function \(f\) as it approaches \(a\) is the limit

\[\lim_{x \to a^+} f(x) = L. \]

The left-side limit of a function \(f\) is

\[\lim_{x \to a^-} f(x) = L. \]

The notation "\(x \to a^-\)" indicates that we only consider values of \(x\) that are less than \(a\) when evaluating the limit. Likewise, for "\(x \to a^+,\)" we consider only values greater than \(a\). One-sided limits are important when evaluating limits containing absolute values \(|x|\), sign \(\text{sgn}(x)\) , floor functions \(\lfloor x \rfloor\), and other piecewise functions.

The image above demonstrates both left- and right-sided limits on a连续函数\(f(x).\)

Find the left- and right-side limits of the signum function \(\text{sgn}(x)\) as \(x \to 0:\)

\[\text{sgn}(x)= \begin{cases} \frac{|x|}{x} && x\neq 0 \\ 0 && x = 0. \end{cases}\]

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Consider the following graph:

From this we see \(\displaystyle \lim_{x \to 0^+} \text{sgn}(x) = 1 \) and \(\displaystyle \lim_{x \to 0^-}\text{sgn}(x) = -1.\ _\square \)

Determine the limit \( \lim\limits_{x \to 1^{-}} \frac{\sqrt{2x}(x-1)}{|x-1|}. \)

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Note that, for \(x<1,\) \(\left | x-1\right |\) can be written as \(-(x-1)\). Hence, the limit is \(\lim\limits_{x \to 1^{-}} \frac{\sqrt{2x}(x-1)}{-(x-1)} = -\sqrt{2}.\ _\square\)

By definition, a two-sided limit

\[\lim_{x \to a} f(x) = L\]

exists if the片面的limits\(\displaystyle \lim_{x \to a^+} f(x)\) and \(\displaystyle \lim_{x \to a^-} f(x)\) are the same.

Compute the limit

\[ \lim_{x \to 1} \frac{|x - 1|}{x - 1} . \]

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Since the absolute value function \(f(x) = |x| \) is defined in a piecewise manner, we have to consider two limits: \(\lim\limits_{x \to 1^+} \frac{|x - 1|}{x - 1} \) and \(\lim\limits_{x \to 1^-} \frac{|x - 1|}{x - 1}. \)

Start with the limit \(\lim\limits_{x \to 1^+} \frac{|x - 1|}{x - 1}.\) For \(x>1,\) \( |x - 1| = x -1. \) So

\[\lim_{x \to 1^+} \frac{|x - 1|}{x - 1} =\lim_{x \to 1^+} \frac{x - 1}{x - 1} =1.\]

Let us now consider the left-hand limit

\[\lim_{x \to 1^-} \frac{|x - 1|}{x - 1}. \]

For \(x<1,\) \(x - 1 = -|x-1|.\) So

\[\lim_{x \to 1^-} \frac{|x-1|}{-|x - 1|} = -1 . \]

So the two-sided limit \( \lim\limits_{x \to 1} \frac{|x - 1|}{x - 1}\) does not exist. \(_\square\)

The image below is a graph of a function \(f(x)\). As shown, it is continuous for all points except \(x = -1\) and \(x=2\) which are its asymptotes. Find all the integer points \(-4

image

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Since the graph is continuous at all points except \(x=-1\) and \(x=2\), the two-sided limit exists at \(x=-3\), \(x=-2\), \(x=0\), \(x=1,\) and \(x=3\). At \(x=2,\) there is no finite value for either of the two-sided limits, since the function increases without bound as the \(x\)-coordinate approaches \(2\) (but see the next section for a further discussion). The situation is similar for \(x=-1.\) So the points \(x=-3\), \(x=-2\), \(x=0\), \(x=1,\) and \(x=3\) are all the integers on which two-sided limits are defined. \(_\square\)