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Introduction

Let there be a real-valued function

\[ f(x) = (1+x)^{\frac{1}{x}}. \]

Clearly, the function is not defined at \(x = 0\).

But we may evaluate \(f(x)\) for values of \(x\) near \(0\) but not equal to zero. Let us put values near \(0\), starting with \(x = 1\) and successively reducing \(x\) by a factor of 10, making the value smaller with each step. Using a calculator, the results are tabulated below:

\[ \begin{array}{|c|c|} \hline x & f(x) = (1+x)^{\frac{1}{x}} \\ \hline 1 & 2 \\ 0.1 & 2.5937424601 \\ 0.01 & 2.704813829421 \\ 0.001 & 2.716923932235 \\ 0.0001 & 2.718145926825 \\ 0.00001 & 2.718268237174 \\ 0.000001 & 2.718280469319 \\ 0.0000001 & 2.718281692544 \\ 0.00000001 & 2.7182818184867 \\ 0.000000001 & 2.718281827099 \\ \hline \end{array} \]

Observations:

What do you observe? We are trying to approach \(x = 0\) (i.e., \(1 > 0.1 > 0.01 > 0.001 > \dots > 0\)) without actually reaching \(0\). Each step reduces \(x\) to a value 10 times smaller than the previous one. Even if we repeat this process infinitely, the value of \(x\) will never equal \(0\). This is because dividing by zero is undefined.

At the same time, observe the corresponding values of \(f(x)\). As \(x\) gets closer to \(0\), the values of \(f(x)\) approach a specific value: approximately \(2.7182818\ldots\).

As \(x\) gets closer to \(0\), the decimal places in \(f(x)\) stabilize and freeze one by one. This specific number, \(2.7182818\ldots\), is known as the exponential constant or Euler's number (\(e\)). It is an irrational number with an infinite, non-repeating decimal expansion.

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What if we move from the left side?

Let us move in steps towards \(0\) as shown in the table:

\[ \begin{array}{|c|c|} \hline x & f(x) = (1+x)^{\frac{1}{x}} \\ \hline -0.1 & 2.8679719907 \\ -0.01 & 2.7319990264 \\ -0.001 & 2.7196422164 \\ -0.0001 & 2.7184177550 \\ -0.00001 & 2.7182954199 \\ -0.000001 & 2.7182831876 \\ -0.0000001 & 2.7182819643 \\ \hline \end{array} \]

When we move from the left side towards \(0\), the value of the function moves closer to the constant \(e\). We say that the limit of the function at \(x = 0\) is the exponential constant.

Consider the function \(f(x) = x^x\), where \(x > 0\). This function is not defined at \(x = 0\), but it is well-defined for all \(x > 0\), that is, on the right side of \(0\). To explore the behavior of this function as \(x\) approaches \(0\) from the right, we calculate \(f(x)\) for progressively smaller positive values of \(x\). The table below summarizes these values:

\[ \begin{array}{|c|c|} \hline x & f(x) = x^x \\ \hline 0.1 & 0.7943282347242815 \\ 0.05 & 0.8608919580232658 \\ 0.01 & 0.954992586021436 \\ 0.005 & 0.9738478944193077 \\ 0.001 & 0.9931160484209338 \\ 0.0005 & 0.9965457582448796 \\ 0.0001 & 0.9989500024812888 \\ 0.00005 & 0.9994750206144874 \\ 0.00001 & 0.9998942459923283 \\ 0.000005 & 0.9999471224534271 \\ \hline \end{array} \]

As we approach \(x = 0\) from the right, the function values \(f(x) = x^x\) get closer and closer to \(1\). This suggests that the function \(f(x)\) converges to a fixed value, even though it is not defined at \(x = 0\).

This numerical exploration helps us understand the function's behavior and the value it approaches, despite being undefined at \(x = 0\).

Limit Points

In this chapter, we will focus on analyzing the behavior of functions at limit points of their domain. But what exactly is a limit point?

Informally, a point \(a\) is called a limit point of a set or interval if we can move closer and closer to \(a\), however close we want, while staying inside the set or interval.For example, \(2\) is a limit point of the open interval \((2, 6)\). We can move as close to \(2\) as we want by choosing points from the interval, such as the sequence:

\[ x_1 = 2.1, \, x_2 = 2.01, \, x_3 = 2.001, \, x_4 = 2.0001, \dots \]

In this sequence, the points \(x_n\) get closer and closer to \(2\), but always remain within the interval \((2, 6)\).

Similarly, \(6\) is a limit point of \((2, 6)\). We can approach \(6\) using a sequence like:

\[ x_1 = 5.9, \, x_2 = 5.99, \, x_3 = 5.999, \, x_4 = 5.9999, \dots \]

Here, the points \(x_n\) get closer and closer to \(6\), while staying inside the interval \((2, 6)\).

The point \(3\) is also a limit point of \((2, 6)\) because we can move closer to \(3\) from within the interval using a sequence like:

\[ x_1 = 2.9, \, x_2 = 2.99, \, x_3 = 2.999, \, \dots \quad \text{or} \quad x_1 = 3.1, \, x_2 = 3.01, \, x_3 = 3.001, \, \dots \]

These sequences approach \(3\) from either side, always staying inside \((2, 6)\).

However, \(1\) is not a limit point of \((2, 6)\) because there are no points in the interval \((2, 6)\) that can get arbitrarily close to \(1\). The interval simply does not "crowd around" \(1\), so it cannot be approached from within \((2, 6)\).

Actually all points from 2 to 6, including 2 and 6, are limit points of \((2, 6)\).

Why Are limit Points Important?

We are not interested in just any point of the domain when studying the behavior of a function \(f(x)\). Instead, we focus on limit points because these are the points where we can approach \(a\) from within the domain of the function, even if \(a\) itself is not part of the domain.

For example, consider the function \(f(x) = \ln x\), which has the domain \((0, \infty)\).

All points in \([0, \infty)\) are limit points of the domain: - For any point \(a > 0\), we can approach \(a\) from within the domain \((0, \infty)\). For instance, we can get closer and closer to \(a = 2\) using a sequence like \(x_1 = 1.9, \, x_2 = 1.99, \, x_3 = 1.999, \dots\), which lies entirely in the domain. - The point \(0\) is also a limit point, even though it is not part of the domain. We can approach \(0\) using a sequence like \(x_1 = 0.1, \, x_2 = 0.01, \, x_3 = 0.001, \dots\), which stays within the domain \((0, \infty)\).

However, the point \(-1\) is not a limit point of the domain. There are no points in the domain \((0, \infty)\) that can get arbitrarily close to \(-1\), as all points in the domain are strictly positive. Thus, \(-1\) is irrelevant for our analysis of the behavior of \(f(x)\).

We are only interested in points within \([0, \infty)\), the set of all limit points of the domain \((0, \infty)\), as these are the points that determine how the function \(f(x) = \ln x\) behaves.

This means that we study the limits of functions only at the limit points of their domain. For any point \(a\) that is not a limit point of the domain, there is no meaningful way to calculate the limit because we cannot approach \(a\) from within the domain.

Think of a limit point as a destination on a path. Even if the destination itself is inaccessible (not part of the domain), we can still analyze what happens as we move closer and closer to it. Limits provide a tool to study the behavior of functions near such points, which is why they are defined at limit points of the domain only.

Definition of a Limit at a Point

Let \(f: A \to \mathbb{R}\), where \(A\) is the domain of \(f\). Let \(a\) be a limit point of \(A\). We define the limit of \(f(x)\) as \(x\) approaches \(a\) as follows:

If \(f(x)\) gets closer and closer to a number \(L\) as \(x\) moves closer and closer to \(a\) (but not equal to \(a\)), then we say that the limit of \(f(x)\) as \(x \to a\) is \(L\), and we write:

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

How to Approach a point a?

When analyzing the behavior of a function \(f(x)\) at a limiting point \(a\), the idea of approaching \(a\) is fundamental. To approach \(a\), we consider values of \(x\) that get closer and closer to \(a\) from within the domain of the function. The way \(x\) approaches \(a\) can vary, but the limit of \(f(x)\) at \(a\) will always depend on the behavior of \(f(x)\) as \(x\) gets arbitrarily close to \(a\), regardless of the path (or sequence) taken.

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\(f(x) = \frac{\sin(x)}{x}\) and \(x \to 0\)

The function \(f(x) = \frac{\sin(x)}{x}\) has the domain \((-\infty, 0) \cup (0, \infty)\), meaning it is not defined at \(x = 0\), but \(0\) is a limiting point of the domain. To analyze the limit \(\lim_{x \to 0} \frac{\sin(x)}{x}\), we consider several sequences of points \(x_n\) approaching \(0\).

Sequence 1: \(x = 0.1, 0.01, 0.001, 0.0001, \dots\)

\[ \begin{array}{|c|c|} \hline x & f(x) = \frac{\sin(x)}{x} \\ \hline 0.1 & 0.998334166468 \\ 0.01 & 0.999983333417 \\ 0.001 & 0.999999833333 \\ 0.0001 & 0.999999998333 \\ 0.00001 & 0.999999999983 \\ \hline \end{array} \]

As \(x \to 0\) from the right (\(x > 0\)), \(f(x)\) approaches \(1\).

Sequence 2: \(x = -0.1, -0.01, -0.001, -0.0001, \dots\)

\[ \begin{array}{|c|c|} \hline x & f(x) = \frac{\sin(x)}{x} \\ \hline -0.1 & 0.998334166468 \\ -0.01 & 0.999983333417 \\ -0.001 & 0.999999833333 \\ -0.0001 & 0.999999998333 \\ -0.00001 & 0.999999999983 \\ \hline \end{array} \]

As \(x \to 0\) from the left (\(x < 0\)), \(f(x)\) also approaches \(1\).

Sequence 3: \(x = 0.1, -0.05, 0.001, -0.0005, 0.00001, \dots\)

\[ \begin{array}{|c|c|} \hline x & f(x) = \frac{\sin(x)}{x} \\ \hline 0.1 & 0.998334166468 \\ -0.05 & 0.999583385414 \\ 0.001 & 0.999999833333 \\ -0.0005 & 0.999999958333 \\ 0.00001 & 0.999999999983 \\ \hline \end{array} \]

In this case, \(x\) alternates between positive and negative values, but as \(x\) gets closer to \(0\), \(f(x)\) still approaches \(1\).

Sequence 4: \(x = \frac{\pi}{10}, \frac{\pi}{20}, \frac{\pi}{30}, \frac{\pi}{40}, \dots\)

\[ \begin{array}{|c|c|} \hline x & f(x) = \frac{\sin(x)}{x} \\ \hline \frac{\pi}{10} & 0.983631643083 \\ \frac{\pi}{20} & 0.995884840246 \\ \frac{\pi}{30} & 0.997659539588 \\ \frac{\pi}{40} & 0.998564034237 \\ \frac{\pi}{50} & 0.999032925199 \\ \hline \end{array} \]

Using multiples of \(\pi\), \(x\) approaches \(0\) from positive values, and \(f(x)\) again approaches \(1\).

From all the sequences, we observe that:

  • As \(x \to 0\), regardless of how \(x\) approaches \(0\), the values of \(f(x) = \frac{\sin(x)}{x}\) approach \(1\).
  • This demonstrates that the limit \(\lim_{x \to 0} \frac{\sin(x)}{x} = 1\), independent of the sequence used to approach \(0\).

The key idea is that when \(x\) approaches \(a\) (here \(a = 0\)), the specific path or sequence taken does not matter. As long as \(x\) gets arbitrarily close to \(a\) within the domain of \(f(x)\), the limit is determined by the behavior of \(f(x)\). This is why we can confidently conclude that:

\[ \lim_{x \to 0} \frac{\sin(x)}{x} = 1. \]

We can also define limits using the idea of sequences.

Limit of a Function Using Sequences

The limit of a function \(f(x)\) at a point \(a\) can be defined using sequences as follows:

Let \(f: A \to \mathbb{R}\), where \(A\) is the domain of \(f(x)\), and let \(a\) be a limiting point of \(A\). We say that the limit of \(f(x)\) as \(x\) approaches \(a\) is \(L\) if:

For any sequence of points \(x_1, x_2, x_3, \dots\) in the domain of \(f(x)\) that gets closer and closer to \(a\) (but \(x_n \neq a\) for all \(n\)), the sequence of function values \(f(x_1), f(x_2), f(x_3), \dots\) approaches \(L\).

In simpler terms:

  • Take any sequence of points \(x_1, x_2, x_3, \dots\) that gets closer to \(a\).
  • If the corresponding function values \(f(x_1), f(x_2), f(x_3), \dots\) get closer and closer to \(L\), then we write:
\[ \lim_{x \to a} f(x) = L. \]

One-Sided Limits

The concept of a one-sided limit focuses on the behavior of a function \(f(x)\) as \(x\) approaches a point \(a\) from only one side—either from the left (\(x \to a^-\)) or from the right (\(x \to a^+\)).

Let \(f: A \to \mathbb{R}\), where \(A\) is the domain of \(f(x)\), and let \(a\) be a limiting point of \(A\).

Right-Hand Limit (\(x \to a^+\)):

The right-hand limit of \(f(x)\) at \(a\) is the value \(L\) that \(f(x)\) approaches as \(x\) gets arbitrarily close to \(a\) from the right (\(x > a\)). Mathematically:

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

In terms of sequences, if \(x_1, x_2, x_3, \dots\) is a sequence of points from the domain \(A\) that gets closer and closer to \(a\) (but always \(x_n > a\), meaning \(x_n\) comes from the right side of \(a\)), and the function values \(f(x_1), f(x_2), f(x_3), \dots\) get closer and closer to some fixed value \(L\), then we say:

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

This means the function values \(f(x)\) approach \(L\) as \(x\) moves toward \(a\) from the right side.

Left-Hand Limit (\(x \to a^-\)):

The left-hand limit of \(f(x)\) at \(a\) is the value \(L\) that \(f(x)\) approaches as \(x\) gets arbitrarily close to \(a\) from the left (\(x < a\)). Mathematically:

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

In terms of sequences, if \(x_1, x_2, x_3, \dots\) is a sequence of points from the domain \(A\) that gets closer and closer to \(a\) (but always \(x_n < a\), meaning \(x_n\) comes from the left side of \(a\)), and the function values \(f(x_1), f(x_2), f(x_3), \dots\) get closer and closer to some fixed value \(L\), then we say:

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

This means the function values \(f(x)\) approach \(L\) as \(x\) moves toward \(a\) from the left side.

Existence of a Limit

The limit of a function \(f(x)\) as \(x \to a\) exists if \(f(x) \to L \in \mathbb{R}\), no matter how we approach \(a\). This means that, regardless of the sequence of points \(x_1, x_2, x_3, \dots\) getting closer and closer to \(a\), the corresponding function values \(f(x_1), f(x_2), f(x_3), \dots\) must approach the same number \(L\).

When Does a Limit Not Exist?

The limit does not exist in the following scenarios:

  1. Function Values Approach Infinity or Negative Infinity:

    If \(f(x)\) becomes arbitrarily large (\(+\infty\)) or arbitrarily small (\(-\infty\)) as \(x\) approaches \(a\), then the limit does not exist because \(f(x)\) does not approach a specific number \(L \in \mathbb{R}\).

  2. Function Oscillates:

    If \(f(x)\) oscillates between different values without settling on a particular number as \(x\) approaches \(a\), the limit does not exist. For example, functions like \(\sin\left(\frac{1}{x}\right)\) near \(x = 0\) show this behavior.

  3. Two Different Sequences Approach Different Values:

    If we can find two sequences of points, both getting arbitrarily close to \(a\), but the corresponding function values approach two different numbers \(L_1 \neq L_2\), the limit does not exist.

  4. One-Sided Limits are Not Equal:

    For most limits, if the left-hand limit (\(\lim_{x \to a^-} f(x)\)) and the right-hand limit (\(\lim_{x \to a^+} f(x)\)) exist but are not equal, the two-sided limit \(\lim_{x \to a} f(x)\) does not exist.

More We Can Learn From Examples

To better understand the existence (or non-existence) of limits, it is helpful to analyze specific examples of functions where these scenarios occur. Examples will clarify how functions behave as \(x\) approaches a particular point \(a\).

Example

Consider the piecewise-defined function:

\[ f(x) = \begin{cases} x & \text{if } x \geq 1, \\ 1 - x & \text{if } x < 1. \end{cases} \]

We want to evaluate \(\lim_{x \to 1} f(x)\). Let us move \(x\) toward \(1\) using the sequence \(x = 1.1, 0.9, 1.01, 0.99, 1.001, 0.999, 1.0001, 0.9999\). The corresponding values of \(f(x)\) are as follows:

\[ \begin{array}{|c|c|} \hline x & f(x) \\ \hline 1.1 & 1.1 \\ 0.9 & 0.1 \\ 1.01 & 1.01 \\ 0.99 & 0.01 \\ 1.001 & 1.001 \\ 0.999 & 0.001 \\ 1.0001 & 1.0001 \\ 0.9999 & 0.0001 \\ \hline \end{array} \]

We observe that,

As \(x\) approaches \(1\):

  • When \(x > 1\), the function values \(f(x) = x\) approach \(1\).
  • When \(x < 1\), the function values \(f(x) = 1 - x\) approach \(0\).

Thus, the values of \(f(x)\) oscillate between \(1\) (when \(x > 1\)) and \(0\) (when \(x < 1\)), depending on whether \(x\) approaches \(1\) from the right (\(x \to 1^+\)) or the left (\(x \to 1^-\)).

Since the function values do not settle on a single number as \(x\) approaches \(1\), the two-sided limit does not exist:

\[ \lim_{x \to 1} f(x) \text{ does not exist}. \]

This example illustrates how oscillating behavior prevents the existence of a limit.

Alternatively, to determine whether \(\lim_{x \to 1} f(x)\) exists, we evaluate the one-sided limits.

Right-Hand Limit (\(x \to 1^+\)):

Let \(x\) approach \(1\) from the right (\(x > 1\)) using the sequence \(x = 1.1, 1.01, 1.001, 1.0001, 1.00001, \dots\). For \(x \geq 1\), the function is defined as \(f(x) = x\). The corresponding function values are:

\[ \begin{array}{|c|c|} \hline x & f(x) \\ \hline 1.1 & 1.1 \\ 1.01 & 1.01 \\ 1.001 & 1.001 \\ 1.0001 & 1.0001 \\ 1.00001 & 1.00001 \\ \hline \end{array} \]

As \(x \to 1^+\), the values of \(f(x)\) approach \(1\). Hence:

\[ \lim_{x \to 1^+} f(x) = 1. \]

Left-Hand Limit (\(x \to 1^-\)):

Let \(x\) approach \(1\) from the left (\(x < 1\)) using the sequence \(x = 0.9, 0.99, 0.999, 0.9999, \dots\). For \(x < 1\), the function is defined as \(f(x) = 1 - x\). The corresponding function values are:

\[ \begin{array}{|c|c|} \hline x & f(x) \\ \hline 0.9 & 0.1 \\ 0.99 & 0.01 \\ 0.999 & 0.001 \\ 0.9999 & 0.0001 \\ \hline \end{array} \]

As \(x \to 1^-\), the values of \(f(x)\) approach \(0\). Hence:

\[ \lim_{x \to 1^-} f(x) = 0. \]

We conclude, the right-hand and left-hand limits are not equal:

\[ \lim_{x \to 1^+} f(x) = 1 \quad \text{and} \quad \lim_{x \to 1^-} f(x) = 0. \]

Since \(\lim_{x \to 1^+} \neq \lim_{x \to 1^-}\), the two-sided limit \(\lim_{x \to 1} f(x)\) does not exist. This shows that the limit fails to exist because the function behaves differently on either side of \(x = 1\).

Understanding the inexistence of limit from the graph of f

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Looking at the graph of the piecewise function:

\[ f(x) = \begin{cases} x & \text{if } x \geq 1, \\ 1 - x & \text{if } x < 1, \end{cases} \]

we observe the following:

  1. Behavior as \(x \to 1^+\) (from the right):

    • For \(x \geq 1\), \(f(x) = x\), which is represented by the blue line.
    • As \(x\) approaches \(1\) from the right (\(x > 1\)), the values of \(f(x)\) approach \(1\).
    • This is marked on the graph by the filled blue dot at \((1, 1)\).
  2. Behavior as \(x \to 1^-\) (from the left):

    • For \(x < 1\), \(f(x) = 1 - x\), which is represented by the red line.
    • As \(x\) approaches \(1\) from the left (\(x < 1\)), the values of \(f(x)\) approach \(0\).
    • This is marked on the graph by the open red circle at \((1, 0)\), indicating that the value of \(f(x)\) does not include this point for \(x < 1\).

Why the Limit Does Not Exist

For a limit to exist at \(x = 1\), the following condition must hold:

\[ \lim_{x \to 1^-} f(x) = \lim_{x \to 1^+} f(x). \]

From the graph:

  • \(\lim_{x \to 1^-} f(x) = 0\) (red line approaching \(1 - x\)).
  • \(\lim_{x \to 1^+} f(x) = 1\) (blue line approaching \(x\)).

Since the left-hand limit (\(0\)) and the right-hand limit (\(1\)) are not equal, the two-sided limit does not exist at \(x = 1\).

Example

Let us show that \(\lim_{x \to 0^+} \sin\left(\frac{1}{x}\right)\) does not exist

We analyze the behavior of the function \(f(x) = \sin\left(\frac{1}{x}\right)\) as \(x\) approaches \(0\) from the right (\(x \to 0^+\)) to determine whether the limit exists.

Consider the sequence \(x = \frac{1}{\pi}, \frac{1}{2\pi}, \frac{1}{3\pi}, \frac{1}{4\pi}, \dots\), which approaches \(0\) from the right. For this sequence, we calculate the corresponding function values:

\[ \begin{array}{|c|c|} \hline x & f(x) = \sin\left(\frac{1}{x}\right) \\ \hline \frac{1}{\pi} & 0 \\ \frac{1}{2\pi} & 0 \\ \frac{1}{3\pi} & 0 \\ \frac{1}{4\pi} & 0 \\ \frac{1}{5\pi} & 0 \\ \hline \end{array} \]

As \(x\) approaches \(0^+\) through this sequence, \(f(x)\) approaches \(0\).

Now consider the sequence \(x = \frac{2}{\pi}, \frac{2}{5\pi}, \frac{2}{9\pi}, \frac{2}{13\pi}, \dots\), which also approaches \(0\) from the right. For this sequence, we calculate the corresponding function values:

\[ \begin{array}{|c|c|} \hline x & f(x) = \sin\left(\frac{1}{x}\right) \\ \hline \frac{2}{\pi} & 1 \\ \frac{2}{5\pi} & 1 \\ \frac{2}{9\pi} & 1 \\ \frac{2}{13\pi} & 1 \\ \frac{2}{17\pi} & 1 \\ \hline \end{array} \]

As \(x\) approaches \(0^+\) through this sequence, \(f(x)\) approaches \(1\).

From these two sequences, we see that as \(x \to 0^+\), \(f(x)\) approaches \(0\) for the first sequence and \(1\) for the second sequence. Since \(f(x)\) approaches different values depending on how \(x\) approaches \(0^+\), the limit \(\lim_{x \to 0^+} \sin\left(\frac{1}{x}\right)\) does not exist.

Alternatively, consider the sequence \(x = \frac{2}{\pi}, \frac{2}{3\pi}, \frac{2}{5\pi}, \frac{2}{7\pi}, \dots\), which also approaches \(0^+\). For this sequence, we calculate the corresponding function values as:

\[ \begin{array}{|c|c|} \hline x & f(x) = \sin\left(\frac{1}{x}\right) \\ \hline \frac{2}{\pi} & 1 \\ \frac{2}{3\pi} & -1 \\ \frac{2}{5\pi} & 1 \\ \frac{2}{7\pi} & -1 \\ \frac{2}{9\pi} & 1 \\ \hline \end{array} \]

As \(x\) approaches \(0^+\) through this sequence, \(f(x)\) oscillates between \(1\) and \(-1\), without settling on a specific value.

Thus, using the first two sequences, we conclude that the limit does not exist because \(f(x)\) approaches different values (\(0\) and \(1\)). Alternatively, using the third sequence, we observe that \(f(x)\) oscillates between \(1\) and \(-1\), further confirming that \(\lim_{x \to 0^+} \sin\left(\frac{1}{x}\right)\) does not exist.

Similarly we can show that (\lim_{x \to 0^-} \sin\left(\frac{1}{x}\right)) also does not exist.

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The graph of \(f(x) = \sin\left(\frac{1}{x}\right)\) illustrates why the limit at \(x = 0\) does not exist. As \(x\) approaches \(0\) from either side, the function oscillates infinitely between \(1\) and \(-1\). These oscillations become increasingly rapid as \(x\) gets closer to \(0\), as shown by the tightly packed waves near the origin. For a limit to exist, the function values must converge to a single value, but here, the function does not settle on any specific number. Instead, it continues to alternate between \(1\) and \(-1\) indefinitely. Hence, the limit \(\lim_{x \to 0} \sin\left(\frac{1}{x}\right)\) does not exist.

Necessary and Sufficient Condition for the Existence of Limits

The limit of a function \(f(x)\) as \(x \to a\) exists if the following three conditions are satisfied:

  1. Left-Hand Limit (LHL) Exists:

    \[ \lim_{x \to a^-} f(x) = L_1 \]

    This means that as \(x\) approaches \(a\) from the left (\(x < a\)), the function values \(f(x)\) approach a specific value \(L_1 \in \mathbb{R}\).

  2. Right-Hand Limit (RHL) Exists:

    \[ \lim_{x \to a^+} f(x) = L_2 \]

    This means that as \(x\) approaches \(a\) from the right (\(x > a\)), the function values \(f(x)\) approach a specific value \(L_2 \in \mathbb{R}\).

  3. Equality of LHL and RHL:

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

    That is, the values of the left-hand limit and right-hand limit must both exist and be equal to the same finite value \(L\).

If these three conditions are satisfied, the two-sided limit exists and is equal to \(L\):

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