Evaluating the Limit of a Function with the Greatest Integer Function

We are asked 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 left side, denoted as \(\mathop {\lim }\limits_{{\rm{x}} \to {0^ - }} {\rm{f}}\left( {\rm{x}} \right)\). The function involves the greatest integer function, \(\left[ {\rm{x}} \right]\), also known as the floor function.

Understanding the Greatest Integer Function as x approaches 0 from the Left

The greatest integer function \(\left[ {\rm{x}} \right]\) gives the largest integer less than or equal to \({\rm{x}}\). When \({\rm{x}}\) approaches \(0\) from the left (i.e., \({\rm{x}} \to {0^ - }\)), \({\rm{x}}\) takes values that are slightly less than \(0\). For example, \({\rm{x}}\) could be \(-0.1\), \(-0.01\), \(-0.001\), and so on.

• For \({\rm{x}} = -0.1\), \(\left[ {\rm{x}} \right] = \left[ {-0.1} \right] = -1\).
• For \({\rm{x}} = -0.01\), \(\left[ {\rm{x}} \right] = \left[ {-0.01} \right] = -1\).
• For any value of \({\rm{x}}\) such that \(-1 \le {\rm{x}} < 0\), the value of \(\left[ {\rm{x}} \right]\) is \(-1\).

Since we are considering the limit as \({\rm{x}} \to {0^ - }\), \({\rm{x}}\) will eventually be in the interval \([-1, 0)\). Therefore, as \({\rm{x}} \to {0^ - }\), the value of \(\left[ {\rm{x}} \right]\) is constant and equal to \(-1\).

Substituting into the Function Expression

Now let's consider the term \(\left[ {\rm{x}} \right] + {\rm{x}}\) in the function \({\rm{f}}\left( {\rm{x}} \right)\) as \({\rm{x}} \to {0^ - }\). Since \(\left[ {\rm{x}} \right] = -1\) for \({\rm{x}}\) approaching \(0\) from the left, the expression becomes:

\(\left[ {\rm{x}} \right] + {\rm{x}} = -1 + {\rm{x}}\)

As \({\rm{x}} \to {0^ - }\), the value of \(-1 + {\rm{x}}\) approaches \(-1 + 0 = -1\). Let's denote \({\rm{y}} = \left[ {\rm{x}} \right] + {\rm{x}}\). So, as \({\rm{x}} \to {0^ - }\), \({\rm{y}} \to -1\).

Rewriting and Evaluating the Limit

The limit we need to evaluate is \(\mathop {\lim }\limits_{{\rm{x}} \to {0^ - }} \frac{{{{\rm{a}}^{\left[ {\rm{x}} \right] + {\rm{x}}}} - 1}}{{\left[ {\rm{x}} \right] + {\rm{x}}}}\). By substituting \({\rm{y}} = \left[ {\rm{x}} \right] + {\rm{x}}\) and considering that \({\rm{y}} \to -1\) as \({\rm{x}} \to {0^ - }\), the limit can be rewritten as:

\(\mathop {\lim }\limits_{{\rm{y}} \to - 1} \frac{{{{\rm{a}}^{\rm{y}}} - 1}}{{\rm{y}}}\)

This is a limit where the variable \({\rm{y}}\) approaches \(-1\). The denominator approaches \(-1\), which is not zero. We can evaluate this limit by direct substitution:

\(\frac{{{{\rm{a}}^{-1}} - 1}}{{-1}}\)

Now, we simplify this expression:

\(\frac{{{{\rm{a}}^{-1}} - 1}}{{-1}} = \frac{{\frac{1}{{\rm{a}}} - 1}}{{-1}}\)

Combine the terms in the numerator:

\(\frac{{\frac{{1 - {\rm{a}}}}{{\rm{a}}}}}{{-1}}\)

Divide the numerator by the denominator:

\(\frac{{1 - {\rm{a}}}}{{\rm{a}}} \times \frac{1}{{-1}} = - \frac{{1 - {\rm{a}}}}{{\rm{a}}}\)

Distribute the negative sign:

\(- \frac{{1 - {\rm{a}}}}{{\rm{a}}} = \frac{{ - 1 + {\rm{a}}}}{{\rm{a}}} = \frac{{{\rm{a}} - 1}}{{\rm{a}}}\)

Separate the terms:

\(\frac{{{\rm{a}} - 1}}{{\rm{a}}} = \frac{{\rm{a}}}{{\rm{a}}} - \frac{1}{{\rm{a}}} = 1 - \frac{1}{{\rm{a}}}\)

Using negative exponents, \(\frac{1}{{\rm{a}}} = {{\rm{a}}^{-1}}\). So the result is \(1 - {{\rm{a}}^{-1}}\).

Conclusion

The limit of the function \({\rm{f}}\left( {\rm{x}} \right)\) as \({\rm{x}}\) approaches \(0\) from the left is \(1 - {{\rm{a}}^{-1}}\).