Question

Let $f(x) = \begin{cases} 2^{1/x}, & x \neq 0 \\ 0, & x = 0 \end{cases}$ Then,

• A. $\lim_{x \to 0} f(x)$ does not exist
• B. $f$ is continuous at $x = 0$
• C. left hand limit at 0 does not exist
• D. $\lim_{x \to 0} f(x) = 0$

Hint:

Whenever $1/x$ appears in an exponent of a base greater than 1 (like $2^{1/x}$), the limit at $0$ always diverges to infinity from the positive direction, making the overall limit non-existent.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
For a limit of a function to exist at a point, both its Left-Hand Limit (LHL) and Right-Hand Limit (RHL) at that point must exist, be finite, and be equal to each other.

Step 2: Detailed Explanation:
Let us evaluate the limits of $f(x)$ as $x$ approaches 0:
1. Right-Hand Limit (RHL) as $x \to 0^+$:
Let $x = 0 + h$ where $h > 0$ and $h \to 0$:
\[ \text{RHL} = \lim_{h \to 0} 2^{1/h} \]
As $h \to 0$, the exponent $\frac{1}{h} \to \infty$:
\[ \text{RHL} = 2^{\infty} = \infty \]
Since the RHL is infinite, it does not exist as a finite real value.
2. Left-Hand Limit (LHL) as $x \to 0^-$:
Let $x = 0 - h$ where $h > 0$ and $h \to 0$:
\[ \text{LHL} = \lim_{h \to 0} 2^{-1/h} \]
As $h \to 0$, the exponent $-\frac{1}{h} \to -\infty$:
\[ \text{LHL} = 2^{-\infty} = \frac{1}{2^{\infty}} = 0 \]
The LHL is equal to 0, which exists and is finite.
Because the RHL tends to infinity and does not equal the LHL, the limit of $f(x)$ as $x$ approaches 0 does not exist.

Step 3: Final Answer
The correct option is (A).