Analyzing Function Continuity at x = 0

We are given the function \(f(x) = \dfrac{1}{1+2^{1/x}}\) and asked to determine its behavior at \(x = 0\).

For a function to be continuous at a point \(x=a\), three conditions must be satisfied:

1. \(f(a)\) must be defined.
2. The limit \(\displaystyle \lim_{x\rightarrow a} f(x)\) must exist.
3. The limit must be equal to the function value, i.e., \(\displaystyle \lim_{x\rightarrow a} f(x) = f(a)\).

Let's check these conditions for our function \(f(x)\) at \(x = 0\).

Step 1: Check if \(f(0)\) is defined

The function is given by \(f(x) = \dfrac{1}{1+2^{1/x}}\). At \(x = 0\), the term \(1/x\) is undefined. Consequently, \(2^{1/x}\) is undefined, and thus \(f(0)\) is undefined.

Since \(f(0)\) is not defined, the first condition for continuity is not met. Therefore, the function is discontinuous at \(x=0\). However, let's examine the limits to understand the nature of the discontinuity, as the options specify reasons related to limits.

Step 2: Evaluate the Left-Hand Limit (LHL) at x = 0

The left-hand limit is the limit as \(x\) approaches \(0\) from the negative side (\(x < 0\)).

\(\displaystyle LHL = \lim_{x\rightarrow 0^-} f(x) = \lim_{x\rightarrow 0^-} \dfrac{1}{1+2^{1/x}}\)

As \(x \rightarrow 0^-\) (meaning \(x\) is a small negative number), the term \(1/x \rightarrow -\infty\). For example, if \(x = -0.1\), \(1/x = -10\); if \(x = -0.001\), \(1/x = -1000\).

So, as \(1/x \rightarrow -\infty\), the term \(2^{1/x} \rightarrow 2^{-\infty}\).

Recall that \(a^{-\infty} = 1/a^{\infty}\) for \(a > 1\). Here \(a=2\), so \(2^{-\infty} = \dfrac{1}{2^\infty} = \dfrac{1}{\infty} = 0\).

Substituting this back into the limit expression:

\(\displaystyle LHL = \dfrac{1}{1+0} = 1\)

So, the left-hand limit at \(x=0\) is 1.

Step 3: Evaluate the Right-Hand Limit (RHL) at x = 0

The right-hand limit is the limit as \(x\) approaches \(0\) from the positive side (\(x > 0\)).

\(\displaystyle RHL = \lim_{x\rightarrow 0^+} f(x) = \lim_{x\rightarrow 0^+} \dfrac{1}{1+2^{1/x}}\)

As \(x \rightarrow 0^+\) (meaning \(x\) is a small positive number), the term \(1/x \rightarrow +\infty\). For example, if \(x = 0.1\), \(1/x = 10\); if \(x = 0.001\), \(1/x = 1000\).

So, as \(1/x \rightarrow +\infty\), the term \(2^{1/x} \rightarrow 2^{+\infty}\).

For \(a > 1\), \(a^{+\infty} = \infty\). Here \(a=2\), so \(2^{+\infty} = \infty\).

Substituting this back into the limit expression:

\(\displaystyle RHL = \dfrac{1}{1+\infty} = \dfrac{1}{\infty} = 0\)

So, the right-hand limit at \(x=0\) is 0.

Step 4: Compare LHL and RHL

We found that \(LHL = 1\) and \(RHL = 0\). Since the left-hand limit and the right-hand limit are not equal (\(1 \ne 0\)), the overall limit \(\displaystyle \lim_{x\rightarrow 0} f(x)\) does not exist.

Conclusion on Discontinuity

A function is discontinuous at a point if any of the three conditions for continuity fail. In this case, \(f(0)\) is undefined, and the limit \(\displaystyle \lim_{x\rightarrow 0} f(x)\) does not exist because the LHL and RHL are different.

The discontinuity arises specifically because the left-hand limit is not equal to the right-hand limit at \(x=0\).

Let's look at the options again based on our analysis:

• Discontinuous because \(L \lim_{x\rightarrow0}f(x){\ne}R\displaystyle\lim_{x\rightarrow0}f(x)\) - This matches our finding that LHL=1 and RHL=0.
• Discontinuous because \(\displaystyle \lim_{x\rightarrow0}f(0){\ne}f(0)\) - The notation \(\displaystyle \lim_{x\rightarrow0}f(0)\) is incorrect. It should be \(\displaystyle \lim_{x\rightarrow0}f(x)\). Also, \(\displaystyle \lim_{x\rightarrow0}f(x)\) does not exist, and \(f(0)\) is undefined, so they cannot be equal. While true that the limit (which doesn't exist) cannot equal the function value (which is undefined), the primary reason for the limit not existing is the unequal LHL and RHL.
• Continuous - This is incorrect.
• Discontinuous because \(R\displaystyle\lim_{x\rightarrow0}f(x)\) does not exist - This is incorrect; RHL exists and is equal to 0.

Therefore, the most accurate reason provided for the discontinuity based on the options is that the left-hand limit does not equal the right-hand limit.

Summary of Findings

At \(x=0\), the function \(f(x) = \dfrac{1}{1+2^{1/x}}\) is undefined. Furthermore, the left-hand limit (1) and the right-hand limit (0) as \(x\) approaches 0 are not equal. This inequality of one-sided limits means the overall limit at \(x=0\) does not exist, causing the function to be discontinuous at \(x=0\).

The discontinuity arises because the function approaches different values from the left side and the right side of \(x=0\).

Revision Table: Function Continuity & Discontinuity

Additional Information: Understanding Limits and Discontinuity

Understanding limits from the left and right sides of a point is crucial for analyzing function behavior, especially at points where the function definition might change or involve terms like \(1/x\). The expression \(2^{1/x}\) behaves very differently as \(x\) approaches 0 from the negative side versus the positive side because the exponent \(1/x\) tends towards negative infinity in one case and positive infinity in the other.

When \(\displaystyle \lim_{x\rightarrow a^-} f(x) \ne \displaystyle \lim_{x\rightarrow a^+} f(x)\), the overall limit \(\displaystyle \lim_{x\rightarrow a} f(x)\) does not exist. This is a common cause of discontinuity, known as a jump discontinuity if both one-sided limits are finite but different.

In this specific function, the term \(2^{1/x}\) is the source of the dramatic change in behavior around \(x=0\). The base 2 is greater than 1. For bases greater than 1, \(a^y \rightarrow \infty\) as \(y \rightarrow \infty\) and \(a^y \rightarrow 0\) as \(y \rightarrow -\infty\). This explains why \(2^{1/x}\) goes to 0 on the left side of 0 and to infinity on the right side.