Understanding the Limit Problem

The problem asks us to find the limit of the function \({\rm{f}}\left( {\rm{x}} \right) = \frac{{{\rm{sin}}\left( {{{\rm{e}}^{{\rm{x}} - 2}} - 1} \right)}}{{{\rm{In}}\left( {{\rm{x}} - 1} \right)}}\) as \({\rm{x}}\) approaches 2. Evaluating limits is a fundamental concept in calculus used to understand the behavior of functions near specific points.

Evaluating the Limit Form

First, let's substitute \({\rm{x}} = 2\) into the function to determine the form of the limit:

• Numerator: \({\rm{sin}}\left( {{{\rm{e}}^{2 - 2}} - 1} \right) = {\rm{sin}}\left( {{{\rm{e}}^0} - 1} \right) = {\rm{sin}}\left( {1 - 1} \right) = {\rm{sin}}\left( 0 \right) = 0\)
• Denominator: \({\rm{In}}\left( {2 - 1} \right) = {\rm{In}}\left( 1 \right) = 0\)

Since we get the form \(\frac{0}{0}\), this is an indeterminate form, which means we need to use other methods, such as standard limit formulas or L'Hopital's Rule, to evaluate the limit.

Applying Standard Limit Formulas

We can simplify this limit using standard limit formulas. Let's make a substitution to make it easier. Let \({\rm{y}} = {\rm{x}} - 2\). As \({\rm{x}} \to 2\), \({\rm{y}} \to 0\). Note that \({\rm{x}} - 1 = \left( {\rm{y}} + 2 \right) - 1 = {\rm{y}} + 1\).

The original limit becomes:

\(\mathop {\lim }\limits_{{\rm{x}} \to 2} \frac{{{\rm{sin}}\left( {{{\rm{e}}^{{\rm{x}} - 2}} - 1} \right)}}{{{\rm{In}}\left( {{\rm{x}} - 1} \right)}} = \mathop {\lim }\limits_{{\rm{y}} \to 0} \frac{{{\rm{sin}}\left( {{{\rm{e}}^{\rm{y}} - 1} \right)}}{{{\rm{In}}\left( {1 + {\rm{y}}} \right)}}\)

We will use the following standard limit formulas:

• \(\mathop {\lim }\limits_{{\rm{z}} \to 0} \frac{{\rm{sin}}\left( {\rm{z}} \right)}{{\rm{z}}} = 1\)
• \(\mathop {\lim }\limits_{{\rm{z}} \to 0} \frac{{{\rm{e}}^{\rm{z}} - 1}}{{\rm{z}}} = 1\)
• \(\mathop {\lim }\limits_{{\rm{z}} \to 0} \frac{{\rm{In}}\left( {1 + {\rm{z}} \right)}{{\rm{z}}} = 1\)

Step-by-Step Calculation using Standard Limits

We can rewrite the expression by multiplying and dividing by terms that help us use the standard formulas:

\(\mathop {\lim }\limits_{{\rm{y}} \to 0} \frac{{{\rm{sin}}\left( {{{\rm{e}}^{\rm{y}} - 1} \right)}}{{{\rm{In}}\left( {1 + {\rm{y}}} \right)}} = \mathop {\lim }\limits_{{\rm{y}} \to 0} \frac{{\frac{{\rm{sin}}\left( {{{\rm{e}}^{\rm{y}} - 1} \right)}{{{{\rm{e}}^{\rm{y}} - 1}}} \times \left( {{{\rm{e}}^{\rm{y}} - 1} \right)}}{{\frac{{\rm{In}}\left( {1 + {\rm{y}}} \right)}{{\rm{y}}} \times {\rm{y}}}}\)

Rearranging the terms:

\(= \mathop {\lim }\limits_{{\rm{y}} \to 0} \left( \frac{{\rm{sin}}\left( {{{\rm{e}}^{\rm{y}} - 1} \right)}{{{{\rm{e}}^{\rm{y}} - 1}}} \times \frac{{{\rm{e}}^{\rm{y}} - 1}}{{\rm{y}}} \times \frac{{\rm{y}}}{{{\rm{In}}\left( {1 + {\rm{y}}} \right)}} \right)\)

Now, let's evaluate each part of the product as \({\rm{y}} \to 0\):

• For the first part, let \({\rm{z}} = {{\rm{e}}^{\rm{y}} - 1}\). As \({\rm{y}} \to 0\), \({\rm{z}} \to {{\rm{e}}^0} - 1 = 1 - 1 = 0\). So, \(\mathop {\lim }\limits_{{\rm{y}} \to 0} \frac{{\rm{sin}}\left( {{{\rm{e}}^{\rm{y}} - 1} \right)}{{{{\rm{e}}^{\rm{y}} - 1}}} = \mathop {\lim }\limits_{{\rm{z}} \to 0} \frac{{\rm{sin}}\left( {\rm{z}} \right)}{{\rm{z}}} = 1\).
• For the second part, this is a direct standard limit: \(\mathop {\lim }\limits_{{\rm{y}} \to 0} \frac{{{\rm{e}}^{\rm{y}} - 1}}{{\rm{y}}} = 1\).
• For the third part, this is the reciprocal of a standard limit: \(\mathop {\lim }\limits_{{\rm{y}} \to 0} \frac{{\rm{y}}}{{{\rm{In}}\left( {1 + {\rm{y}}} \right)}} = \frac{1}{{\mathop {\lim }\limits_{{\rm{y}} \to 0} \frac{{\rm{In}}\left( {1 + {\rm{y}}} \right)}{{\rm{y}}}}} = \frac{1}{1} = 1\).

Multiplying the results of each part:

\(\mathop {\lim }\limits_{{\rm{y}} \to 0} \left( \frac{{\rm{sin}}\left( {{{\rm{e}}^{\rm{y}} - 1} \right)}{{{{\rm{e}}^{\rm{y}} - 1}}} \times \frac{{{\rm{e}}^{\rm{y}} - 1}}{{\rm{y}}} \times \frac{{\rm{y}}}{{{\rm{In}}\left( {1 + {\rm{y}}} \right)}} \right) = 1 \times 1 \times 1 = 1\)

Alternative Method: L'Hopital's Rule

Since the limit is in the \(\frac{0}{0}\) indeterminate form as \({\rm{x}} \to 2\), we can apply L'Hopital's Rule. L'Hopital's Rule states that if \(\mathop {\lim }\limits_{{\rm{x}} \to {\rm{c}}} \frac{{\rm{f}}\left( {\rm{x}} \right)}{{\rm{g}}\left( {\rm{x}} \right)}\) is of the form \(\frac{0}{0}\) or \(\frac{\infty}{\infty}\), then \(\mathop {\lim }\limits_{{\rm{x}} \to {\rm{c}}} \frac{{\rm{f}}\left( {\rm{x}} \right)}{{\rm{g}}\left( {\rm{x}} \right)} = \mathop {\lim }\limits_{{\rm{x}} \to {\rm{c}}} \frac{{{\rm{f}}'\left( {\rm{x}} \right)}}{{{\rm{g}}'\left( {\rm{x}} \right)}}\), provided the latter limit exists.

Let \({\rm{f}}\left( {\rm{x}} \right) = {\rm{sin}}\left( {{{\rm{e}}^{{\rm{x}} - 2}} - 1} \right)\) and \({\rm{g}}\left( {\rm{x}} \right) = {\rm{In}}\left( {{\rm{x}} - 1} \right)\).

Find the derivatives \({\rm{f}}'\left( {\rm{x}} \right)\) and \({\rm{g}}'\left( {\rm{x}} \right)\):

• \({\rm{f}}'\left( {\rm{x}} \right) = \frac{{\rm{d}}}{{{\rm{dx}}}}\left( {\rm{sin}}\left( {{{\rm{e}}^{{\rm{x}} - 2}} - 1} \right) \right)\)
  Using the chain rule, \(\frac{{\rm{d}}}{{{\rm{du}}}}\left( {\rm{sin}}\left( {\rm{u}} \right) \right) = {\rm{cos}}\left( {\rm{u}} \right)\) and \(\frac{{\rm{d}}}{{{\rm{dx}}}}\left( {{{\rm{e}}^{{\rm{x}} - 2}} - 1} \right) = {{\rm{e}}^{{\rm{x}} - 2}} \times \frac{{\rm{d}}}{{{\rm{dx}}}}\left( {{\rm{x}} - 2} \right) - 0 = {{\rm{e}}^{{\rm{x}} - 2}}\).
  So, \({\rm{f}}'\left( {\rm{x}} \right) = {\rm{cos}}\left( {{{\rm{e}}^{{\rm{x}} - 2}} - 1} \right) \times {{\rm{e}}^{{\rm{x}} - 2}}\).
• \({\rm{g}}'\left( {\rm{x}} \right) = \frac{{\rm{d}}}{{{\rm{dx}}}}\left( {\rm{In}}\left( {{\rm{x}} - 1} \right) \right)\)
  Using the chain rule, \(\frac{{\rm{d}}}{{{\rm{du}}}}\left( {\rm{In}}\left( {\rm{u}} \right) \right) = \frac{1}{{\rm{u}}}\) and \(\frac{{\rm{d}}}{{{\rm{dx}}}}\left( {{\rm{x}} - 1} \right) = 1\).
  So, \({\rm{g}}'\left( {\rm{x}} \right) = \frac{1}{{{\rm{x}} - 1}} \times 1 = \frac{1}{{{\rm{x}} - 1}}\).

Now apply L'Hopital's Rule:

\(\mathop {\lim }\limits_{{\rm{x}} \to 2} \frac{{{\rm{f}}'\left( {\rm{x}} \right)}}{{{\rm{g}}'\left( {\rm{x}} \right)}} = \mathop {\lim }\limits_{{\rm{x}} \to 2} \frac{{{{\rm{e}}^{{\rm{x}} - 2}} {\rm{cos}}\left( {{{\rm{e}}^{{\rm{x}} - 2}} - 1} \right)}}{\frac{1}{{{\rm{x}} - 1}}}\)

Substitute \({\rm{x}} = 2\):

\(= \frac{{{{\rm{e}}^{2 - 2}} {\rm{cos}}\left( {{{\rm{e}}^{2 - 2}} - 1} \right)}}{\frac{1}{{2 - 1}}} = \frac{{{{\rm{e}}^0} {\rm{cos}}\left( {{{\rm{e}}^0} - 1} \right)}}{\frac{1}{1}} = \frac{{1 \times {\rm{cos}}\left( {1 - 1} \right)}}{1} = \frac{{\rm{cos}}\left( 0 \right)}{1} = \frac{1}{1} = 1\)

Both methods yield the same result, 1.

Conclusion

The limit of the given function as \({\rm{x}}\) approaches 2 is 1.

Final Answer is 1.

Revision Table: Limit Evaluation Concepts

Additional Information: Standard Limits and Applications

Standard limits are crucial tools for evaluating more complex limits, especially when they result in indeterminate forms. They often arise from the definition of the derivative or Taylor series expansions of functions like sine, exponential, and logarithm around 0.

• The limit \(\mathop {\lim }\limits_{{\rm{x}} \to 0} \frac{{\rm{sin}}\left( {\rm{x}} \right)}{{\rm{x}}} = 1\) is fundamental in calculus, particularly for finding derivatives of trigonometric functions.
• The limit \(\mathop {\lim }\limits_{{\rm{x}} \to 0} \frac{{{\rm{e}}^{\rm{x}} - 1}}{{\rm{x}}} = 1\) is closely related to the derivative of \({\rm{e}}^{\rm{x}}\) at \({\rm{x}} = 0\).
• The limit \(\mathop {\lim }\limits_{{\rm{x}} \to 0} \frac{{\rm{In}}\left( {1 + {\rm{x}}} \right)}{{\rm{x}}} = 1\) is related to the derivative of \({\rm{In}}\left( {\rm{x}} \right)\) at \({\rm{x}} = 1\).

Recognizing when and how to manipulate a limit expression to fit these standard forms is a key skill. This often involves algebraic manipulation, substitution, and splitting the limit into simpler parts if possible. While L'Hopital's Rule is powerful, using standard limits is often quicker if the expression can be easily transformed.