Key Concepts and Formulas

• Continuity at a Point: A function $f(x)$ is continuous at $x=c$ if and only if $\lim_{x \to c} f(x) = f(c)$. This implies that the limit exists at $x=c$, and its value is equal to the function's value at $c$.
• Taylor Series Expansion: For functions like $e^y$ around $y=0$, the Taylor series expansion is useful for evaluating limits of indeterminate forms. The expansion for $e^y$ is $e^y = 1 + y + \frac{y^2}{2!} + \frac{y^3}{3!} + \dots$.
• L'Hôpital's Rule (Implicitly used via series expansion): While not explicitly stated in the provided solution, the concept of ensuring the lowest power of $x$ in the numerator and denominator cancel out is related to L'Hôpital's rule when dealing with $\frac{0}{0}$ indeterminate forms. Taylor series provide a more direct way to handle this in this specific problem.

Step-by-Step Solution

Step 1: Understand the Condition for Continuity The problem states that the function $f(x)$ is continuous at $x=0$. For continuity at $x=0$, the following condition must hold: $\lim_{x \to 0} f(x) = f(0)$ This means we first need to find the limit of $f(x)$ as $x$ approaches $0$.

Step 2: Set up the Limit Expression Substitute the given definition of $f(x)$ into the limit expression: $\lim_{x \to 0} f(x) = \lim_{x \to 0} \left( \frac{1}{x} - \frac{k - 1}{e^{2x} - 1} \right)$

Step 3: Combine the Terms to Form a Single Fraction To evaluate this limit, we need to combine the two terms into a single fraction by finding a common denominator, which is $x(e^{2x} - 1)$: $\lim_{x \to 0} \left( \frac{e^{2x} - 1 - x(k - 1)}{x(e^{2x} - 1)} \right) = \lim_{x \to 0} \left( \frac{e^{2x} - 1 - kx + x}{x(e^{2x} - 1)} \right)$

Step 4: Use Taylor Series Expansion for $e^{2x}$ As $x \to 0$, the expression is of the indeterminate form $\frac{0}{0}$. To simplify, we use the Taylor series expansion of $e^y$ around $y=0$, which is $e^y = 1 + y + \frac{y^2}{2!} + \frac{y^3}{3!} + \dots$. Substituting $y = 2x$, we get: $e^{2x} = 1 + (2x) + \frac{(2x)^2}{2!} + \frac{(2x)^3}{3!} + \dots = 1 + 2x + \frac{4x^2}{2} + \frac{8x^3}{6} + \dots$ Now, substitute this expansion into the numerator and the denominator of our limit expression.

Step 5: Expand and Simplify the Numerator Let's expand the numerator: $e^{2x} - 1 - kx + x$. $\text{Numerator} = \left( 1 + 2x + \frac{4x^2}{2!} + \frac{8x^3}{3!} + \dots \right) - 1 - kx + x$ $\text{Numerator} = (1 - 1) + (2x + x - kx) + \frac{4x^2}{2!} + \frac{8x^3}{3!} + \dots$ $\text{Numerator} = (3 - k)x + \frac{4x^2}{2!} + \frac{8x^3}{3!} + \dots$

Step 6: Expand and Simplify the Denominator Now, let's expand the denominator: $x(e^{2x} - 1)$. $\text{Denominator} = x \left( \left( 1 + 2x + \frac{4x^2}{2!} + \frac{8x^3}{3!} + \dots \right) - 1 \right)$ $\text{Denominator} = x \left( 2x + \frac{4x^2}{2!} + \frac{8x^3}{3!} + \dots \right)$ $\text{Denominator} = 2x^2 + \frac{4x^3}{2!} + \frac{8x^4}{3!} + \dots$

Step 7: Rewrite the Limit with Expanded Terms Now, substitute the simplified numerator and denominator back into the limit expression: $\lim_{x \to 0} \left( \frac{(3 - k)x + \frac{4x^2}{2!} + \frac{8x^3}{3!} + \dots}{2x^2 + \frac{4x^3}{2!} + \frac{8x^4}{3!} + \dots} \right)$

Step 8: Analyze the Limit for Existence For the limit to exist, the lowest power of $x$ in the numerator must be greater than or equal to the lowest power of $x$ in the denominator. The lowest power of $x$ in the denominator is $x^2$ (from the term $2x^2$). The lowest power of $x$ in the numerator is $x$ (from the term $(3-k)x$), unless its coefficient is zero.

For the limit to be a finite number, the term with the lowest power of $x$ in the numerator must match or be of a higher power than the lowest power in the denominator. If $(3-k) \neq 0$, the numerator would have an $x$ term, while the denominator starts with $x^2$. This would lead to a limit of $\pm \infty$, which means the function would not be continuous. Therefore, to ensure the limit exists, the coefficient of the lowest power of $x$ in the numerator must be zero. $3 - k = 0$ This implies: $k = 3$

Step 9: Evaluate the Limit with $k=3$ Substitute $k=3$ back into the limit expression. The numerator becomes: $\text{Numerator} = (3 - 3)x + \frac{4x^2}{2!} + \frac{8x^3}{3!} + \dots = 0x + \frac{4x^2}{2!} + \frac{8x^3}{3!} + \dots = \frac{4x^2}{2!} + \frac{8x^3}{3!} + \dots$ The limit now becomes: $\lim_{x \to 0} \left( \frac{\frac{4x^2}{2!} + \frac{8x^3}{3!} + \dots}{2x^2 + \frac{4x^3}{2!} + \dots} \right)$ We can factor out $x^2$ from both the numerator and the denominator: $\lim_{x \to 0} \left( \frac{x^2 \left( \frac{4}{2!} + \frac{8x}{3!} + \dots \right)}{x^2 \left( 2 + \frac{4x}{2!} + \dots \right)} \right)$ Cancel out the $x^2$ terms: $\lim_{x \to 0} \left( \frac{\frac{4}{2!} + \frac{8x}{3!} + \dots}{2 + \frac{4x}{2!} + \dots} \right)$ Now, as $x \to 0$, all terms with $x$ in them will go to zero: $\frac{\frac{4}{2!} + 0 + \dots}{2 + 0 + \dots} = \frac{\frac{4}{2}}{2} = \frac{2}{2} = 1$ So, the limit of $f(x)$ as $x \to 0$ is $1$.

Step 10: Determine $f(0)$ Since the function is continuous at $x=0$, we have $f(0) = \lim_{x \to 0} f(x)$. Therefore, $f(0) = 1$.

Step 11: Form the Ordered Pair (k, f(0)) We found that $k = 3$ and $f(0) = 1$. The ordered pair is $(k, f(0)) = (3, 1)$.

Common Mistakes & Tips

• Incorrect Taylor Series Expansion: Ensure the Taylor series for $e^{2x}$ is correctly written and substituted, paying close attention to the powers of $2x$.
• Algebraic Errors: Combining fractions and simplifying terms can be prone to algebraic mistakes. Double-check each step.
• Understanding the Limit Existence Condition: The crucial step is realizing that for the limit to exist when the denominator's lowest power is $x^2$, the numerator's lowest power must also be at least $x^2$. This is achieved by making the coefficient of the $x$ term in the numerator zero.

Summary

To ensure the function $f(x)$ is continuous at $x=0$, we must have $f(0) = \lim_{x \to 0} f(x)$. We evaluated the limit by combining the terms of $f(x)$ into a single fraction and using the Taylor series expansion of $e^{2x}$. For the limit to exist, the coefficient of the lowest power of $x$ in the numerator must be zero, which allowed us to determine that $k=3$. After substituting $k=3$, we found the limit to be $1$. Therefore, $f(0) = 1$, and the ordered pair $(k, f(0))$ is $(3, 1)$.

The final answer is \boxed{(3, 1)}.