Evaluating Limits with Greatest Integer Function

The problem asks us to find the limit of the function \({\rm{f}}\left( {\rm{x}} \right) = \frac{{{{\rm{a}}^{\left[ {\rm{x}} \right] + {\rm{x}}}} - 1}}{{\left[ {\rm{x}} \right] + {\rm{x}}}}\) as \({\rm{x}}\) approaches 0 from the positive side (denoted as \({\rm{x}} \to {0^ + }\)). The notation \([{\rm{x}}]\) represents the greatest integer function, which gives the largest integer less than or equal to \({\rm{x}}\).

Understanding the Function Near the Limit Point

We are interested in the behavior of the function as \({\rm{x}}\) approaches 0 from values greater than 0. Let's consider values of \({\rm{x}}\) that are very close to 0 but positive, such as 0.001, 0.0001, and so on.

For any positive value of \({\rm{x}}\) such that \(0 < {\rm{x}} < 1\), the greatest integer less than or equal to \({\rm{x}}\) is 0. That is, \([{\rm{x}}] = 0\) for \({\rm{x}} \in (0, 1)\).

Simplifying the Expression for the Limit

As \({\rm{x}} \to {0^ + }\), \({\rm{x}}\) is in the interval \((0, 1)\) for values sufficiently close to 0. Therefore, \([{\rm{x}}] = 0\).

Substituting \([{\rm{x}}] = 0\) into the expression \([{\rm{x}}] + {\rm{x}}\), we get \(0 + {\rm{x}} = {\rm{x}}\).

So, as \({\rm{x}} \to {0^ + }\), the function \({\rm{f}}\left( {\rm{x}} \right)\) behaves like:

$$ {\rm{f}}\left( {\rm{x}} \right) = \frac{{{{\rm{a}}^{\left[ {\rm{x}} \right] + {\rm{x}}}} - 1}}{{\left[ {\rm{x}} \right] + {\rm{x}}}} \to \frac{{{{\rm{a}}^{\rm{x}}} - 1}}{{\rm{x}}} \quad \text{as } {\rm{x}} \to {0^ + } $$

Now, the problem reduces to finding the limit of \(\frac{{{{\rm{a}}^{\rm{x}}} - 1}}{{\rm{x}}}\) as \({\rm{x}} \to 0\).

Applying Standard Limits

This is a standard limit form. The limit of \(\frac{{{{\rm{a}}^{\rm{x}}} - 1}}{{\rm{x}}}\) as \({\rm{x}} \to 0\) is known to be \(\ln({\rm{a}})\), where \(\ln\) denotes the natural logarithm (logarithm to the base \(e\)).

The formula is: \(\mathop {\lim }\limits_{{\rm{x}} \to 0} \frac{{{{\rm{a}}^{\rm{x}}} - 1}}{{\rm{x}}} = \ln({\rm{a}})\).

Since we are evaluating the limit as \({\rm{x}} \to {0^ + }\), and the expression simplifies to \(\frac{{{{\rm{a}}^{\rm{x}}} - 1}}{{\rm{x}}}\) for \({\rm{x}} > 0\) close to 0, we can apply this standard limit result directly.

Calculating the Limit

Using the standard limit formula, we have:

$$ \mathop {\lim }\limits_{{\rm{x}} \to {0^ + }} {\rm{f}}\left( {\rm{x}} \right) = \mathop {\lim }\limits_{{\rm{x}} \to {0^ + }} \frac{{{{\rm{a}}^{\left[ {\rm{x}} \right] + {\rm{x}}}} - 1}}{{\left[ {\rm{x}} \right] + {\rm{x}}}} $$

As \({\rm{x}} \to {0^ + }\), \([{\rm{x}}] = 0\), so \([{\rm{x}}] + {\rm{x}} = {\rm{x}}\). The limit becomes:

$$ \mathop {\lim }\limits_{{\rm{x}} \to {0^ + }} \frac{{{{\rm{a}}^{\rm{x}}} - 1}}{{\rm{x}}} $$

Applying the standard limit \(\mathop {\lim }\limits_{{\rm{x}} \to 0} \frac{{{{\rm{a}}^{\rm{x}}} - 1}}{{\rm{x}}} = \ln({\rm{a}})\), we get:

$$ \mathop {\lim }\limits_{{\rm{x}} \to {0^ + }} {\rm{f}}\left( {\rm{x}} \right) = \ln({\rm{a}}) $$

The result is \(\ln({\rm{a}})\).

Final Answer

The limit of the function \({\rm{f}}\left( {\rm{x}} \right)\) as \({\rm{x}} \to {0^ + }\) is \(\ln({\rm{a}})\).

Revision Table: Key Concepts for Function Limits

Additional Information: Exploring Limits and Functions

Understanding limits is fundamental in calculus. They describe the behavior of a function as the input approaches a certain value. For functions involving the greatest integer part, the limit often depends on whether the approach is from the left (\({\rm{x}} \to {\rm{c}}^- \)) or the right (\({\rm{x}} \to {\rm{c}}^+ \)), especially if the limit point 'c' is an integer.

• Greatest Integer Function near 0: When \({\rm{x}} \to {0^ + }\), \({\rm{x}}\) is a small positive number, so \([{\rm{x}}] = 0\). When \({\rm{x}} \to {0^- }\), \({\rm{x}}\) is a small negative number (e.g., -0.001), so \([{\rm{x}}] = -1\). This difference is critical for evaluating two-sided limits or left-hand limits. In this problem, only the right-hand limit was required.
• Standard Limits: Certain limit forms appear frequently and have known results. \(\mathop {\lim }\limits_{{\rm{x}} \to 0} \frac{{\sin {\rm{x}}}}{{\rm{x}}} = 1\) and \(\mathop {\lim }\limits_{{\rm{x}} \to 0} \frac{{{{\rm{e}}^{\rm{x}}} - 1}}{{\rm{x}}} = 1\) (which is a special case of \(\frac{{{{\rm{a}}^{\rm{x}}} - 1}}{{\rm{x}}}\) when \({\rm{a}} = {\rm{e}}\), giving \(\ln({\rm{e}}) = 1\)) are other examples. Recognizing these forms simplifies complex limit problems.
• Natural Logarithm: The natural logarithm \(\ln({\rm{a}})\) is the power to which \(e\) must be raised to get \({\rm{a}}\). It naturally appears in limits involving exponential functions, reflecting the relationship between exponentiation and logarithms.

This problem combined the evaluation of a greatest integer function near a point with the application of a standard exponential limit, demonstrating how different concepts interact in calculus problems.