Limits: Introduction and Evaluation Techniques

Concept of a Limit: Definition and Understanding

The concept of a limit is one of the most fundamental building blocks of calculus and mathematical analysis. It provides a way to describe the behavior of a function as the input variable approaches a certain value, without necessarily being equal to that value. Understanding limits is essential for grasping concepts like continuity, derivatives, and integrals.

Intuitive Understanding

Informally, the limit of a function $f(x)$ as $x$ approaches a specific value $a$ is the value that $f(x)$ gets arbitrarily close to as $x$ gets arbitrarily close to $a$, from either side, but without actually reaching $a$. We are interested in the function's behavior *near* the point, not necessarily *at* the point.

Think of it as asking: "If we take input values ($x$) closer and closer to $a$, what output value ($f(x)$) does the function seem to be aiming for?"

Consider the function $f(x) = \frac{x^2 - 1}{x - 1}$. If we try to evaluate $f(x)$ at $x=1$, we get $\frac{1^2 - 1}{1 - 1} = \frac{0}{0}$, which is an indeterminate form, meaning the function is undefined at $x=1$. However, let's see what happens as $x$ gets very close to 1.

For any value of $x$ other than 1 (i.e., $x \neq 1$), we can simplify the function algebraically:

$f(x) = \frac{x^2 - 1}{x - 1} = \frac{(x-1)(x+1)}{x-1}$

Since $x \neq 1$, the term $(x-1)$ is non-zero, allowing us to cancel it from the numerator and denominator:

$f(x) = x+1 \quad \text{for } x \neq 1$.

Now, let's pick values of $x$ that are increasingly close to 1, but not equal to 1:

As the table shows, regardless of whether $x$ approaches 1 from values less than 1 or values greater than 1, the corresponding values of $f(x)$ get closer and closer to 2. Even though $f(1)$ is undefined, the function is clearly heading towards the value 2.

Graphically, the function $f(x) = \frac{x^2-1}{x-1}$ for $x \neq 1$ is the same as the line $y = x+1$ but with a 'hole' at the point $(1, 2)$. The limit describes the height of the graph at that hole.

Notation

We denote the limit of the function $f(x)$ as $x$ approaches $a$ is $L$ using the following notation:

$\lim\limits_{x \to a} f(x) = L$

This is read as "the limit of $f(x)$ as $x$ approaches $a$ equals $L$." The arrow $x \to a$ signifies that $x$ is getting arbitrarily close to $a$, and the equality $\dots = L$ signifies that the corresponding function value $f(x)$ is getting arbitrarily close to $L$. It's important to remember that the limit is about the behavior *near* $a$, not necessarily *at* $a$.

Formal Definition (Epsilon-Delta Definition)

To provide a rigorous, unambiguous definition of a limit, mathematicians use the epsilon-delta definition. This definition precisely captures the intuitive idea that $f(x)$ can be made arbitrarily close to $L$ by taking $x$ sufficiently close to $a$ (but not equal to $a$).

We say that the limit of $f(x)$ as $x$ approaches $a$ is $L$, written as $\lim\limits_{x \to a} f(x) = L$, if for every real number $\epsilon > 0$ (epsilon, representing a desired closeness to $L$), there exists a corresponding real number $\delta > 0$ (delta, representing a required closeness to $a$) such that for all $x$ in the domain of $f$, if the distance between $x$ and $a$ is positive but less than $\delta$, i.e., $0 < |x - a| < \delta$, then the distance between $f(x)$ and $L$ is less than $\epsilon$, i.e., $|f(x) - L| < \epsilon$.

In simpler terms:

• Start with an arbitrary positive value $\epsilon$. This sets a target band $(L-\epsilon, L+\epsilon)$ around $L$ on the y-axis.
• The definition requires that we must be able to find a positive value $\delta$. This $\delta$ defines an interval $(a-\delta, a+\delta)$ around $a$ on the x-axis.
• The condition $0 < |x - a| < \delta$ means we consider all $x$ values within the $\delta$-interval around $a$, *excluding* $a$ itself. This is often called a "punctured neighbourhood" of $a$.
• The condition $|f(x) - L| < \epsilon$ means that for any $x$ chosen in the punctured $\delta$-interval around $a$, its function value $f(x)$ *must* lie within the $\epsilon$-interval around $L$.

So, no matter how small you make the tolerance $\epsilon$ around $L$, you must be able to find a corresponding tolerance $\delta$ around $a$ such that any $x$ (not equal to $a$) within $\delta$ of $a$ maps to an $f(x)$ within $\epsilon$ of $L$. This formalizes the notion of "arbitrarily close." Proving limits using the epsilon-delta definition is a rigorous exercise often covered in introductory analysis courses.

Left Hand Limit and Right Hand Limit: Definition and Significance

When examining the behavior of a function $f(x)$ as $x$ approaches a point $a$, the approach can be made from two directions: from values less than $a$ or from values greater than $a$. The limits from these two directions are called the Left Hand Limit and the Right Hand Limit, respectively. These one-sided limits are crucial for determining the existence of the overall limit and for understanding the continuity of a function.

Left Hand Limit (LHL)

The Left Hand Limit of a function $f(x)$ as $x$ approaches $a$ is the value that $f(x)$ approaches as $x$ gets arbitrarily close to $a$ from values that are less than $a$. We are considering the behavior of the function to the immediate left of $a$ on the x-axis.

Notation: The left hand limit is commonly denoted by adding a superscript minus sign to the value $a$ in the limit notation:

$\lim\limits_{x \to a^-} f(x)$

Other notations sometimes encountered include $\lim\limits_{x \uparrow a} f(x)$ or $f(a-)$ or $f(a-0)$. The superscript minus sign indicates that $x$ approaches $a$ from the left side (i.e., $x < a$).

When evaluating a left hand limit, we consider a sequence of $x$ values like $a - 0.1, a - 0.01, a - 0.001, \dots$, approaching $a$ from smaller values.

Right Hand Limit (RHL)

The Right Hand Limit of a function $f(x)$ as $x$ approaches $a$ is the value that $f(x)$ approaches as $x$ gets arbitrarily close to $a$ from values that are greater than $a$. We are considering the behavior of the function to the immediate right of $a$ on the x-axis.

Notation: The right hand limit is commonly denoted by adding a superscript plus sign to the value $a$ in the limit notation:

$\lim\limits_{x \to a^+} f(x)$

Other notations sometimes encountered include $\lim\limits_{x \downarrow a} f(x)$ or $f(a+)$ or $f(a+0)$. The superscript plus sign indicates that $x$ approaches $a$ from the right side (i.e., $x > a$).

When evaluating a right hand limit, we consider a sequence of $x$ values like $a + 0.1, a + 0.01, a + 0.001, \dots$, approaching $a$ from larger values.

Significance of One-Sided Limits

Left hand and right hand limits are fundamental tools in calculus and have significant importance for several reasons:

• Existence of the Bilateral Limit: The most crucial role of one-sided limits is in determining whether the overall (or bilateral) limit $\lim\limits_{x \to a} f(x)$ exists. The limit $\lim\limits_{x \to a} f(x) = L$ exists if and only if both the left hand limit and the right hand limit at $x=a$ exist and are equal to the same value $L$. That is, $\lim\limits_{x \to a} f(x)$ exists if and only if $\lim\limits_{x \to a^-} f(x) = \lim\limits_{x \to a^+} f(x) = L$. If the LHL and RHL are different, or if either does not exist, then the limit $\lim\limits_{x \to a} f(x)$ does not exist.
• Analysis of Piecewise Functions: Functions defined piecewise (using different formulas for different intervals of $x$) often exhibit different behavior on either side of the points where the definition changes. One-sided limits are essential for analyzing the behavior of such functions at these critical points.
• Definition of Continuity: One-sided limits are directly involved in the definition of continuity at a point. A function $f(x)$ is continuous at $x=a$ if and only if $\lim\limits_{x \to a^-} f(x) = \lim\limits_{x \to a^+} f(x) = f(a)$. This condition requires that the LHL, the RHL, and the function value at the point all exist and are equal.
• Behavior near Discontinuities and Asymptotes: One-sided limits help categorize different types of discontinuities (like jump discontinuities, where LHL $\neq$ RHL) and describe the behavior of functions near vertical asymptotes (where one or both one-sided limits might be $\pm \infty$).

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Existence of a Limit

For the overall limit of a function $f(x)$ as $x$ approaches a point $a$, denoted by $\lim\limits_{x \to a} f(x)$, to exist and have a finite value, the behavior of the function as $x$ approaches $a$ from the left must be the same as its behavior as $x$ approaches $a$ from the right. This requires the agreement of the one-sided limits.

Condition for the Existence of a Finite Limit

The limit of a function $f(x)$ as $x$ approaches $a$, $\lim\limits_{x \to a} f(x)$, exists and is equal to a finite value $L$ if and only if both the left hand limit (LHL) and the right hand limit (RHL) as $x$ approaches $a$ exist and are equal to that same finite value $L$.

Mathematically, this condition can be stated as:

is true if and only if

and

So, if conditions (i), (ii), and (iii) are all met, then the bilateral limit $\lim\limits_{x \to a} f(x)$ exists and is equal to $L$. Conversely, if the limit $\lim\limits_{x \to a} f(x)$ exists and is equal to $L$, it implies that both one-sided limits exist and are equal to $L$.

This fundamental theorem connects the two-sided limit to the one-sided limits and is crucial for evaluating limits of functions, especially piecewise-defined functions.

Cases Where the Limit Does Not Exist (DNE)

The limit $\lim\limits_{x \to a} f(x)$ does not exist (DNE) if the condition for existence (LHL = RHL = finite value) is not met. This can happen in several ways:

1. The Left Hand Limit and the Right Hand Limit Exist but are Not Equal:

   If $\lim\limits_{x \to a^-} f(x) = L_1$ and $\lim\limits_{x \to a^+} f(x) = L_2$, but $L_1 \neq L_2$, then the limit $\lim\limits_{x \to a} f(x)$ does not exist. The function "jumps" at $x=a$.

   Example: Consider a piecewise function like $f(x) = \begin{cases} x+1 & \text{if } x < 0 \\ x-1 & \text{if } x \ge 0 \end{cases}$. As $x \to 0^-$, $f(x) \to 1$. As $x \to 0^+$, $f(x) \to -1$. Since $1 \neq -1$, $\lim\limits_{x \to 0} f(x)$ does not exist.

2. One or Both One-Sided Limits are Unbounded:

   If $\lim\limits_{x \to a^-} f(x) = \pm \infty$ or $\lim\limits_{x \to a^+} f(x) = \pm \infty$ (or both), then the limit $\lim\limits_{x \to a} f(x)$ does not exist as a finite number. While we might describe the behavior by saying the limit is infinity, technically the limit does not exist in the sense of converging to a finite value $L$. This indicates a vertical asymptote at $x=a$.

   Example: Consider $f(x) = \frac{1}{x^2}$ as $x \to 0$. $\lim\limits_{x \to 0^-} \frac{1}{x^2} = +\infty$ and $\lim\limits_{x \to 0^+} \frac{1}{x^2} = $+\infty$. Even though both go to $+\infty$, the limit does not exist as a finite value. For $f(x) = \frac{1}{x}$ as $x \to 0$, $\lim\limits_{x \to 0^-} \frac{1}{x} = -\infty$ and $\lim\limits_{x \to 0^+} \frac{1}{x} = +\infty$. The limits are different and unbounded, so the overall limit does not exist.

3. One or Both One-Sided Limits Do Not Exist Due to Oscillation:

   If the function values oscillate infinitely rapidly as $x$ approaches $a$ and do not approach any single value, then the limit does not exist. This is often the case for functions involving trigonometric expressions near points where their arguments become unbounded.

   Example: Consider $f(x) = \sin\left(\frac{1}{x}\right)$ as $x \to 0$. As $x$ gets closer to 0, $\frac{1}{x}$ gets larger and larger (in magnitude), causing $\sin\left(\frac{1}{x}\right)$ to oscillate infinitely often between -1 and 1. The function values do not settle on a single value, so $\lim\limits_{x \to 0} \sin\left(\frac{1}{x}\right)$ does not exist.

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Evaluation of Limits by Direct Substitution

The simplest and most straightforward method for evaluating limits is through direct substitution. This method relies on the property of continuity of the function at the point the variable is approaching. If a function is "well-behaved" at a certain point, its limit as the variable approaches that point is simply the value of the function at that point.

Principle of Direct Substitution

If a function $f(x)$ is continuous at $x = a$, then the limit of $f(x)$ as $x$ approaches $a$ is equal to the value of the function evaluated at $x=a$.

This can be stated formally as:

If $f$ is continuous at $a$, then $\lim\limits_{x \to a} f(x) = f(a)$.

Recall from the concept of limits and continuity that for a function to be continuous at $x=a$, it must satisfy three conditions: 1) $f(a)$ is defined, 2) $\lim\limits_{x \to a} f(x)$ exists, and 3) $\lim\limits_{x \to a} f(x) = f(a)$. The direct substitution method is essentially applying the third condition of continuity in reverse: if we know the function is continuous, we can use $f(a)$ to find the limit.

Functions for which Direct Substitution is Applicable

Direct substitution works for a wide range of functions at points within their domain where they are continuous. Many common types of functions are continuous on their entire domain or on specific intervals. These include:

• Polynomial Functions: Any polynomial function $P(x) = c_n x^n + c_{n-1} x^{n-1} + \dots + c_1 x + c_0$ is continuous for all real numbers. Therefore, for any polynomial $P(x)$ and any real number $a$:

  $\lim\limits_{x \to a} P(x) = P(a)$

• Rational Functions: A rational function is a ratio of two polynomials, $R(x) = \frac{P(x)}{Q(x)}$. A rational function is continuous everywhere in its domain, which is all real numbers except where the denominator $Q(x)$ is zero. Thus, for a rational function $\frac{P(x)}{Q(x)}$ and a value $a$, direct substitution is valid if $Q(a) \neq 0$:

  If $Q(a) \neq 0$, then $\lim\limits_{x \to a} \frac{P(x)}{Q(x)} = \frac{P(a)}{Q(a)}$.

  If $Q(a) = 0$, direct substitution results in an undefined form, and other methods (like factorization or rationalization) must be used, or the limit may not exist (e.g., a vertical asymptote). The case where $Q(a)=0$ and $P(a)=0$ results in the indeterminate form $\frac{0}{0}$, which will be discussed in the next section.

• Trigonometric Functions: Basic trigonometric functions like $\sin x$ and $\cos x$ are continuous for all real numbers. Functions like $\tan x = \frac{\sin x}{\cos x}$ are continuous on their domain (all real numbers except where $\cos x = 0$, i.e., $x \neq \frac{\pi}{2} + n\pi$, $n \in \mathbb{Z}$). Direct substitution works at any point within the function's domain:

  $\lim\limits_{x \to a} \sin x = \sin a$}

  $\lim\limits_{x \to a} \cos x = \cos a$}

  $\lim\limits_{x \to a} \tan x = \tan a$, provided $\cos a \neq 0$.

• Exponential Functions: The exponential function $b^x$ (where $b>0, b\neq 1$) and $e^x$ are continuous for all real numbers.

  $\lim\limits_{x \to a} b^x = b^a$}

  $\lim\limits_{x \to a} e^x = e^a$}

• Logarithmic Functions: The logarithmic function $\log_b x$ (where $b>0, b\neq 1$) and $\ln x = \log_e x$ are continuous on their domain, which is all positive real numbers ($x > 0$). Direct substitution works for any positive value $a$:

  $\lim\limits_{x \to a} \log_b x = \log_b a$ (for $a > 0$)

  $\lim\limits_{x \to a} \ln x = \ln a$ (for $a > 0$)

• Root Functions: Functions like $\sqrt[n]{x}$ are continuous on their domain. For even roots (e.g., $\sqrt{x}$), the domain is $x \ge 0$, so direct substitution works for $a \ge 0$. For odd roots (e.g., $\sqrt[3]{x}$), the domain is all real numbers, so direct substitution works for any real $a$.

  $\lim\limits_{x \to a} \sqrt{x} = \sqrt{a}$ (for $a \ge 0$)

  $\lim\limits_{x \to a} \sqrt[3]{x} = \sqrt[3]{a}$ (for any real $a$).

• Combinations of Functions: The sum, difference, product, and composition of continuous functions are continuous. The quotient of two continuous functions is continuous wherever the denominator is non-zero. Therefore, direct substitution also works for combinations of the above functions, provided the point of approach $a$ is within the domain of the resulting function and does not make a denominator zero.

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Methods for Evaluating Indeterminate Forms (Factorization, Rationalization)

As we saw with direct substitution, it is the simplest way to evaluate limits for many functions. However, direct substitution cannot be used when it results in an indeterminate form. The most common indeterminate form encountered initially is $\frac{0}{0}$. An indeterminate form means that the limit cannot be determined simply by looking at the form; it requires further analysis and manipulation of the expression. It does not necessarily mean the limit does not exist.

Indeterminate Form $\frac{0}{0}$

If, upon attempting direct substitution of $x=a$ into a rational expression $\frac{f(x)}{g(x)}$, we find that both the numerator $f(a)$ and the denominator $g(a)$ are equal to zero, then the expression is in the indeterminate form $\frac{0}{0}$ at $x=a$.

That is, if $\lim\limits_{x \to a} f(x) = 0$ and $\lim\limits_{x \to a} g(x) = 0$, then $\lim\limits_{x \to a} \frac{f(x)}{g(x)}$ is of the form $\frac{0}{0}$.

When we encounter this indeterminate form, it signifies that there is likely a common factor, often $(x-a)$, in both the numerator and the denominator that is causing both to be zero at $x=a$. By eliminating this common factor through algebraic techniques, we can simplify the expression and then evaluate the limit using direct substitution on the simplified form.

Two primary methods for dealing with the $\frac{0}{0}$ indeterminate form at an elementary level are Factorization and Rationalization.

1. Factorization Method

This method is most effective when the numerator and denominator are polynomial expressions that can be factored. The goal is to identify and cancel the common factor $(x-a)$ that leads to the $\frac{0}{0}$ form.

Steps for the Factorization Method:

1. First, attempt direct substitution of $x=a$ into the expression $\frac{f(x)}{g(x)}$. Confirm that this results in the indeterminate form $\frac{0}{0}$. If it does not, direct substitution was the correct method, or the limit might be $\pm \infty$ or non-existent for other reasons (which factorization won't resolve).
2. Factor the numerator $f(x)$ completely and factor the denominator $g(x)$ completely. Since $f(a)=0$ and $g(a)=0$, the Factor Theorem from algebra tells us that $(x-a)$ must be a factor of both $f(x)$ and $g(x)$.
3. Cancel the common factor $(x-a)$ from the numerator and the denominator. This step is valid because when we are evaluating the limit as $x \to a$, we are considering values of $x$ that are very close to $a$ but are not equal to $a$. Therefore, the factor $(x-a)$ is non-zero, and division by a non-zero quantity is permissible.
4. Once the common factor $(x-a)$ is cancelled, the simplified expression is equivalent to the original function for all $x \neq a$. Evaluate the limit of this simplified expression using direct substitution of $x=a$. This will give the value of the limit.

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2. Rationalization Method

This method is commonly used when the expression involves radicals (usually square roots) and direct substitution results in the $\frac{0}{0}$ indeterminate form. The process involves multiplying the numerator and denominator by the conjugate of the radical expression to eliminate the root, which then usually reveals a common factor that can be cancelled.

The key formula used here is the difference of squares: $(A - B)(A + B) = A^2 - B^2$.

Steps for the Rationalization Method:

1. First, attempt direct substitution of $x=a$ into the expression $\frac{f(x)}{g(x)}$. Confirm that this results in the indeterminate form $\frac{0}{0}$.
2. Identify the term(s) containing the radical(s). These are typically in the form $(\sqrt{A} \pm B)$ or $(A \pm \sqrt{B})$.
3. Multiply both the numerator and the denominator by the conjugate of the radical term. The conjugate of $(\sqrt{A} - B)$ is $(\sqrt{A} + B)$, and the conjugate of $(\sqrt{A} + B)$ is $(\sqrt{A} - B)$. Similarly, the conjugate of $(A - \sqrt{B})$ is $(A + \sqrt{B})$, and the conjugate of $(A + \sqrt{B})$ is $(A - \sqrt{B})$.
4. Simplify the numerator and denominator. Use the difference of squares formula $(u-v)(u+v) = u^2 - v^2$ on the part that was rationalized (e.g., $(\sqrt{A} - B)(\sqrt{A} + B) = (\sqrt{A})^2 - B^2 = A - B^2$). It is often beneficial NOT to fully expand the other part (the one that now contains the conjugate factor) initially, as a common factor is usually about to appear.
5. After simplification, look for the common factor, which will often be $(x-a)$, that was causing the $\frac{0}{0}$ form. Cancel this common factor from the numerator and the denominator. This is valid because $x \neq a$.
6. Evaluate the limit of the resulting simplified expression using direct substitution of $x=a$.

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