Calculating the Limit of \({\rm{e}}^{ - \frac{1}{{{{\rm{x}}^2}}}}\) as \({\rm{x}} \to 0\)

We are asked to find the limit of the function \(f(x) = {{\rm{e}}^{ - \frac{1}{{{{\rm{x}}^2}}}}}\) as \({\rm{x}}\) approaches 0. Let's analyze the behavior of the function as \({\rm{x}}\) gets closer and closer to 0.

The function involves an exponential term \({\rm{e}}^u\), where the exponent is \(u = - \frac{1}{{{{\rm{x}}^2}}}\).

We need to figure out what happens to the exponent \( - \frac{1}{{{{\rm{x}}^2}}}\) as \({\rm{x}} \to 0\).

• As \({\rm{x}} \to 0\), the value of \({\rm{x}}^2\) approaches 0.
• Since \({\rm{x}}\) is squared, \({\rm{x}}^2\) is always positive for any \({\rm{x}} \neq 0\). So, \({\rm{x}}^2\) approaches 0 from the positive side (we write this as \({\rm{x}}^2 \to 0^+\)).
• Now consider \( \frac{1}{{{{\rm{x}}^2}}}\). As the denominator \({\rm{x}}^2\) approaches 0 from the positive side, the fraction \( \frac{1}{{{{\rm{x}}^2}}}\) becomes very large and positive. In limit terms, \( \frac{1}{{{{\rm{x}}^2}}} \to +\infty\) as \({\rm{x}} \to 0\).
• Finally, consider the exponent \( - \frac{1}{{{{\rm{x}}^2}}}\). Since \( \frac{1}{{{{\rm{x}}^2}}} \to +\infty\), the term \( - \frac{1}{{{{\rm{x}}^2}}}\) will approach negative infinity. So, \( - \frac{1}{{{{\rm{x}}^2}}} \to -\infty\) as \({\rm{x}} \to 0\).

Now we can evaluate the limit of the original function. We have the form \({\rm{e}}^u\) where \(u \to -\infty\).

We know the behavior of the exponential function \({\rm{e}}^y\) as \(y\) approaches negative infinity. The graph of \(y = {\rm{e}}^x\) shows that as \(x\) moves towards the left (to \(-\infty\)), the value of \({\rm{e}}^x\) approaches 0.

Therefore, as the exponent \( - \frac{1}{{{{\rm{x}}^2}}}\) approaches \(-\infty\), the function \( {{\rm{e}}^{ - \frac{1}{{{{\rm{x}}^2}}}}}\) approaches \({\rm{e}}^{-\infty}\), which is 0.

Mathematically, we can write this as:

Let \(y = - \frac{1}{{{{\rm{x}}^2}}}\). As \({\rm{x}} \to 0\), \({\rm{x}}^2 \to 0^+\), so \(\frac{1}{{{{\rm{x}}^2}}} \to +\infty\), which means \(y \to -\infty\).

So, the original limit becomes:

\[ \mathop {{\rm{lim}}}\limits_{{\rm{x}} \to 0} {{\rm{e}}^{ - \frac{1}{{{{\rm{x}}^2}}}}} = \mathop {{\rm{lim}}}\limits_{y \to -\infty} {{\rm{e}}^{y}} \]

We know that \( \mathop {{\rm{lim}}}\limits_{y \to -\infty} {{\rm{e}}^{y}} = 0 \).

Thus, the limit of \( {{\rm{e}}^{ - \frac{1}{{{{\rm{x}}^2}}}}}\) as \({\rm{x}}\) approaches 0 is 0.

The final answer is 0.

Revision Table: Key Concepts for Limit of \({\rm{e}}^{ - \frac{1}{{{{\rm{x}}^2}}}}\)

Additional Information on Evaluating Limits

Evaluating limits is a fundamental concept in calculus. When dealing with limits, especially those involving indeterminate forms or specific function types like exponential functions, it's crucial to analyze the behavior of the function as the input approaches the limit point.

• Direct Substitution: First, try substituting the limit value into the function. If you get a defined value, that's usually the limit. However, this doesn't work if the function is undefined at that point (e.g., division by zero).
• Analyzing Function Behavior: For functions like \( \frac{1}{{{{\rm{x}}^2}}}\) or \({\rm{e}}^y\), understanding their graphs and how they behave as the input goes to 0, infinity, or minus infinity is key.
• Left and Right Limits: For some functions, the limit as \({\rm{x}}\) approaches a point might depend on whether \({\rm{x}}\) approaches from the left or the right. In this problem, since \({\rm{x}}^2\) is involved, the sign of \({\rm{x}}\) doesn't affect \({\rm{x}}^2\), so the left and right limits are the same.
• Standard Limits: Knowing common limits, such as \( \mathop {{\rm{lim}}}\limits_{y \to -\infty} {{\rm{e}}^{y}} = 0 \) or \( \mathop {{\rm{lim}}}\limits_{y \to +\infty} {{\rm{e}}^{y}} = +\infty \), is very helpful.

In this specific problem, the key was recognizing that the exponent \( - \frac{1}{{{{\rm{x}}^2}}}\) tends to \(-\infty\) as \({\rm{x}} \to 0\), and then applying the known limit of the exponential function as its argument tends to \(-\infty\).