Key Concepts and Formulas

• Continuity of a Function: A function $f(x)$ is continuous at a point $x=a$ if $\lim_{x \to a} f(x) = f(a)$. To make a function continuous at a point where it is undefined, we need to define its value at that point to be equal to the limit of the function as $x$ approaches that point.
• L'Hôpital's Rule: If $\lim_{x \to c} \frac{g(x)}{h(x)}$ is of the indeterminate form $\frac{0}{0}$ or $\frac{\infty}{\infty}$, then $\lim_{x \to c} \frac{g(x)}{h(x)} = \lim_{x \to c} \frac{g'(x)}{h'(x)}$, provided the latter limit exists.
• Taylor Series Expansion: The Taylor series expansion of $e^u$ around $u=0$ is $e^u = 1 + u + \frac{u^2}{2!} + \frac{u^3}{3!} + \dots$. For small $u$, $e^u \approx 1+u$.

Step-by-Step Solution

Step 1: Understand the Problem and Identify the Goal We are given a function $f(x) = \frac{1}{x} - \frac{2}{e^{2x} - 1}$ defined on $R \setminus \{0\}$. We need to find the value of $f(0)$ that makes the function continuous at $x=0$. For continuity at $x=0$, we must have $f(0) = \lim_{x \to 0} f(x)$.

Step 2: Rewrite the Function to Combine Terms To evaluate the limit, it's helpful to combine the two terms into a single fraction. $f(x) = \frac{1}{x} - \frac{2}{e^{2x} - 1}$ $f(x) = \frac{(e^{2x} - 1) - 2x}{x(e^{2x} - 1)}$

Step 3: Evaluate the Limit of the Combined Function Now, we need to find the limit of this combined expression as $x \to 0$. $\lim_{x \to 0} f(x) = \lim_{x \to 0} \frac{e^{2x} - 1 - 2x}{x(e^{2x} - 1)}$ Let's substitute $x=0$ into the numerator and the denominator to check the form of the limit. Numerator: $e^{2(0)} - 1 - 2(0) = e^0 - 1 - 0 = 1 - 1 = 0$. Denominator: $0(e^{2(0)} - 1) = 0(1 - 1) = 0(0) = 0$. Since the limit is in the indeterminate form $\frac{0}{0}$, we can apply L'Hôpital's Rule.

Step 4: Apply L'Hôpital's Rule (First Application) Let $g(x) = e^{2x} - 1 - 2x$ and $h(x) = x(e^{2x} - 1) = xe^{2x} - x$. We need to find the derivatives of $g(x)$ and $h(x)$. $g'(x) = \frac{d}{dx}(e^{2x} - 1 - 2x) = 2e^{2x} - 2$. $h'(x) = \frac{d}{dx}(xe^{2x} - x) = (1 \cdot e^{2x} + x \cdot 2e^{2x}) - 1 = e^{2x} + 2xe^{2x} - 1$.

Now, apply L'Hôpital's Rule: $\lim_{x \to 0} \frac{g'(x)}{h'(x)} = \lim_{x \to 0} \frac{2e^{2x} - 2}{e^{2x} + 2xe^{2x} - 1}$ Let's substitute $x=0$ again into this new expression. Numerator: $2e^{2(0)} - 2 = 2(1) - 2 = 0$. Denominator: $e^{2(0)} + 2(0)e^{2(0)} - 1 = 1 + 0 - 1 = 0$. The limit is still in the indeterminate form $\frac{0}{0}$, so we need to apply L'Hôpital's Rule again.

Step 5: Apply L'Hôpital's Rule (Second Application) Let $g_1(x) = 2e^{2x} - 2$ and $h_1(x) = e^{2x} + 2xe^{2x} - 1$. We need to find the derivatives of $g_1(x)$ and $h_1(x)$. $g_1'(x) = \frac{d}{dx}(2e^{2x} - 2) = 2(2e^{2x}) = 4e^{2x}$. $h_1'(x) = \frac{d}{dx}(e^{2x} + 2xe^{2x} - 1) = 2e^{2x} + (2 \cdot e^{2x} + 2x \cdot 2e^{2x}) - 0 = 2e^{2x} + 2e^{2x} + 4xe^{2x} = 4e^{2x} + 4xe^{2x}$.

Now, apply L'Hôpital's Rule again: $\lim_{x \to 0} \frac{g_1'(x)}{h_1'(x)} = \lim_{x \to 0} \frac{4e^{2x}}{4e^{2x} + 4xe^{2x}}$ Let's substitute $x=0$ into this expression. Numerator: $4e^{2(0)} = 4(1) = 4$. Denominator: $4e^{2(0)} + 4(0)e^{2(0)} = 4(1) + 0 = 4$. This is no longer an indeterminate form.

Step 6: Calculate the Final Limit Value $\lim_{x \to 0} \frac{4e^{2x}}{4e^{2x} + 4xe^{2x}} = \frac{4}{4} = 1$

Step 7: Define $f(0)$ for Continuity For the function $f(x)$ to be continuous at $x=0$, we must define $f(0)$ to be equal to the limit we just calculated. Therefore, $f(0) = \lim_{x \to 0} f(x) = 1$.

Alternative Method using Taylor Series Expansion

Step 1: Recall Taylor Series for $e^{2x}$ For small $x$, the Taylor series expansion of $e^u$ around $u=0$ is $e^u = 1 + u + \frac{u^2}{2!} + \dots$. Substituting $u=2x$, we get $e^{2x} = 1 + (2x) + \frac{(2x)^2}{2!} + \dots = 1 + 2x + 2x^2 + \dots$.

Step 2: Substitute the Taylor Series into the Function We have $f(x) = \frac{e^{2x} - 1 - 2x}{x(e^{2x} - 1)}$. Substitute the Taylor expansion of $e^{2x}$ into the numerator: $e^{2x} - 1 - 2x = (1 + 2x + 2x^2 + \dots) - 1 - 2x = 2x^2 + \dots$ (terms of order $x^3$ and higher).

Substitute the Taylor expansion of $e^{2x}$ into the denominator: $x(e^{2x} - 1) = x((1 + 2x + 2x^2 + \dots) - 1) = x(2x + 2x^2 + \dots) = 2x^2 + 2x^3 + \dots$.

Step 3: Evaluate the Limit using the Taylor Series Expansion Now, substitute these back into the limit expression: $\lim_{x \to 0} f(x) = \lim_{x \to 0} \frac{2x^2 + \dots}{2x^2 + 2x^3 + \dots}$ Divide both the numerator and the denominator by $2x^2$ (the lowest power of $x$ in the numerator): $\lim_{x \to 0} \frac{1 + \frac{\dots}{2x^2}}{1 + x + \dots}$ As $x \to 0$, the higher order terms go to zero. $\lim_{x \to 0} f(x) = \frac{1}{1} = 1$

Step 4: Define $f(0)$ for Continuity For continuity at $x=0$, $f(0)$ must be equal to this limit. Therefore, $f(0) = 1$.

Common Mistakes & Tips

• Incorrectly applying L'Hôpital's Rule: Ensure that the limit is indeed in an indeterminate form ($\frac{0}{0}$ or $\frac{\infty}{\infty}$) before applying the rule. Applying it otherwise will lead to incorrect results.
• Algebraic Errors: Be meticulous with algebraic manipulations, especially when combining fractions or differentiating complex expressions. A small error can propagate and lead to a wrong answer.
• Taylor Series Approximation: When using Taylor series, make sure to include enough terms to capture the behavior of the function near the point of interest, especially when dealing with subtractions that cancel out leading terms. For this problem, the $x^2$ term was crucial.

Summary

To make the function $f(x) = \frac{1}{x} - \frac{2}{e^{2x} - 1}$ continuous at $x=0$, we need to define $f(0)$ such that it equals the limit of $f(x)$ as $x$ approaches $0$. By combining the terms of $f(x)$ into a single fraction, we obtain $\frac{e^{2x} - 1 - 2x}{x(e^{2x} - 1)}$. This expression results in the indeterminate form $\frac{0}{0}$ as $x \to 0$. We can then apply L'Hôpital's Rule twice, or use the Taylor series expansion of $e^{2x}$, to evaluate the limit. Both methods yield a limit of $1$. Therefore, defining $f(0) = 1$ will make the function continuous at $x=0$.

The final answer is $\boxed{1}$.