Key Concepts and Formulas

• Continuity of a Function: A function $f(x)$ is continuous at a point $x = c$ if the following three conditions are met:

  1. $f(c)$ is defined.
  2. $\mathop {\lim }\limits_{x \to c} f(x)$ exists.
  3. $\mathop {\lim }\limits_{x \to c} f(x) = f(c)$. For a piecewise function defined at $x=c$, this means the Left-Hand Limit (LHL), Right-Hand Limit (RHL), and the function value at $c$ must all be equal: $\mathop {\lim }\limits_{x \to c^-} f(x) = \mathop {\lim }\limits_{x \to c^+} f(x) = f(c)$.

• Standard Limits:

  • $\mathop {\lim }\limits_{\theta \to 0} \frac{\sin \theta}{\theta} = 1$
  • $\mathop {\lim }\limits_{\theta \to 0} \frac{\log_e (1 + \theta)}{\theta} = 1$
  • $\mathop {\lim }\limits_{\theta \to 0} \frac{1 - \cos \theta}{\theta^2} = \frac{1}{2}$

• Trigonometric Identities:

  • $\cos^2 x - \sin^2 x = \cos(2x)$

Step-by-Step Solution

Step 1: Understand the condition for continuity. The problem states that the function $f(x)$ is continuous at $x=0$. This implies that the Left-Hand Limit (LHL) at $x=0$, the Right-Hand Limit (RHL) at $x=0$, and the function value at $x=0$ must all be equal. $\mathop {\lim }\limits_{x \to {0^- }} f(x) = \mathop {\lim }\limits_{x \to {0^+ }} f(x) = f(0)$ From the definition of the function, we have $f(0) = k$.

Step 2: Calculate the Right-Hand Limit (RHL). For $x > 0$, $f(x) = \frac{{{{\cos }^2}x - {{\sin }^2}x - 1}}{{\sqrt {{x^2} + 1} - 1}}$. We use the identity $\cos^2 x - \sin^2 x = \cos(2x)$. $\mathop {\lim }\limits_{x \to {0^+ }} f(x) = \mathop {\lim }\limits_{x \to {0^+ }} \frac{{{{\cos }^2}x - {{\sin }^2}x - 1}}{{\sqrt {{x^2} + 1} - 1}}$ Substitute the identity: $\mathop {\lim }\limits_{x \to {0^+ }} \frac{{\cos(2x) - 1}}{{\sqrt {{x^2} + 1} - 1}}$ This limit is of the indeterminate form $\frac{0}{0}$ as $x \to 0$. To evaluate it, we can use rationalization and standard limits. Multiply the numerator and denominator by the conjugate of the denominator, $(\sqrt{x^2 + 1} + 1)$: $\mathop {\lim }\limits_{x \to {0^+ }} \frac{{\cos(2x) - 1}}{{\sqrt {{x^2} + 1} - 1}} \times \frac{{\sqrt {{x^2} + 1} + 1}}{{\sqrt {{x^2} + 1} + 1}}$ $= \mathop {\lim }\limits_{x \to {0^+ }} \frac{{(\cos(2x) - 1)(\sqrt {{x^2} + 1} + 1)}}{{(x^2 + 1) - 1}}$ $= \mathop {\lim }\limits_{x \to {0^+ }} \frac{{(\cos(2x) - 1)(\sqrt {{x^2} + 1} + 1)}}{{x^2}}$ Rearrange the terms: $= \mathop {\lim }\limits_{x \to {0^+ }} \frac{{\cos(2x) - 1}}{{x^2}} \times \mathop {\lim }\limits_{x \to {0^+ }} (\sqrt {{x^2} + 1} + 1)$ For the first part, $\mathop {\lim }\limits_{x \to {0^+ }} \frac{{\cos(2x) - 1}}{{x^2}}$, we can use the standard limit $\mathop {\lim }\limits_{\theta \to 0} \frac{1 - \cos \theta}{\theta^2} = \frac{1}{2}$. Let $\theta = 2x$. As $x \to 0$, $\theta \to 0$. $\mathop {\lim }\limits_{x \to {0^+ }} \frac{{\cos(2x) - 1}}{{x^2}} = \mathop {\lim }\limits_{x \to {0^+ }} -\frac{{1 - \cos(2x)}}{{x^2}}$ To match the standard form, we need $(2x)^2 = 4x^2$ in the denominator. $= \mathop {\lim }\limits_{x \to {0^+ }} -4 \times \frac{{1 - \cos(2x)}}{{(2x)^2}}$ Using the standard limit, this becomes: $= -4 \times \frac{1}{2} = -2$ Now, evaluate the second part: $\mathop {\lim }\limits_{x \to {0^+ }} (\sqrt {{x^2} + 1} + 1) = \sqrt {0^2 + 1} + 1 = \sqrt{1} + 1 = 1 + 1 = 2$ Combining the two parts: $\mathop {\lim }\limits_{x \to {0^+ }} f(x) = (-2) \times 2 = -4$

Step 3: Calculate the Left-Hand Limit (LHL). For $x < 0$, $f(x) = \frac{1}{x} \log_e \left( \frac{1 + \frac{x}{a}}{1 - \frac{x}{b}} \right)$. $\mathop {\lim }\limits_{x \to {0^- }} f(x) = \mathop {\lim }\limits_{x \to {0^- }} \frac{1}{x} \log_e \left( \frac{1 + \frac{x}{a}}{1 - \frac{x}{b}} \right)$ Using the logarithm property $\log \frac{A}{B} = \log A - \log B$: $\mathop {\lim }\limits_{x \to {0^- }} \frac{1}{x} \left[ \log_e \left( 1 + \frac{x}{a} \right) - \log_e \left( 1 - \frac{x}{b} \right) \right]$ Distribute the $\frac{1}{x}$: $\mathop {\lim }\limits_{x \to {0^- }} \left[ \frac{\log_e \left( 1 + \frac{x}{a} \right)}{x} - \frac{\log_e \left( 1 - \frac{x}{b} \right)}{x} \right]$ We use the standard limit $\mathop {\lim }\limits_{\theta \to 0} \frac{\log_e (1 + \theta)}{\theta} = 1$. For the first term, let $\theta_1 = \frac{x}{a}$. As $x \to 0^-$, $\theta_1 \to 0$. We need $\frac{x}{a}$ in the denominator. $\mathop {\lim }\limits_{x \to {0^- }} \frac{\log_e \left( 1 + \frac{x}{a} \right)}{x} = \mathop {\lim }\limits_{x \to {0^- }} \frac{\log_e \left( 1 + \frac{x}{a} \right)}{\frac{x}{a}} \times \frac{1}{a}$ This limit is $1 \times \frac{1}{a} = \frac{1}{a}$. For the second term, let $\theta_2 = -\frac{x}{b}$. As $x \to 0^-$, $\theta_2 \to 0$. We need $-\frac{x}{b}$ in the denominator. $\mathop {\lim }\limits_{x \to {0^- }} \frac{\log_e \left( 1 - \frac{x}{b} \right)}{x} = \mathop {\lim }\limits_{x \to {0^- }} \frac{\log_e \left( 1 + (-\frac{x}{b}) \right)}{-\frac{x}{b}} \times \left(-\frac{1}{b}\right)$ This limit is $1 \times \left(-\frac{1}{b}\right) = -\frac{1}{b}$. So, the LHL is: $\mathop {\lim }\limits_{x \to {0^- }} f(x) = \frac{1}{a} - \left(-\frac{1}{b}\right) = \frac{1}{a} + \frac{1}{b}$

Step 4: Apply the continuity condition. Since $f(x)$ is continuous at $x=0$, we have LHL = RHL = $f(0)$. From Step 2, RHL = -4. From Step 3, LHL = $\frac{1}{a} + \frac{1}{b}$. From the function definition, $f(0) = k$. Therefore, we have: $\frac{1}{a} + \frac{1}{b} = -4$ And also: $k = -4$

Step 5: Calculate the required expression. We need to find the value of $\frac{1}{a} + \frac{1}{b} + \frac{4}{k}$. We have found that $\frac{1}{a} + \frac{1}{b} = -4$ and $k = -4$. Substitute these values into the expression: $\left( \frac{1}{a} + \frac{1}{b} \right) + \frac{4}{k} = (-4) + \frac{4}{(-4)}$ $= -4 + (-1)$ $= -5$

Common Mistakes & Tips

• Algebraic Errors: Be extremely careful with signs when manipulating fractions, logarithms, and trigonometric terms. For example, $\cos(2x) - 1$ is negative for small $x \neq 0$.
• Standard Limit Application: Ensure the argument of the standard limit is correctly matched in the denominator. For $\frac{\log(1+\theta)}{\theta}$, if the argument is $\frac{x}{a}$, you need $\frac{x}{a}$ in the denominator. Similarly for trigonometric limits.
• Rationalization: When rationalizing a denominator of the form $\sqrt{A} - B$, multiply by $\sqrt{A} + B$. Ensure the conjugate is applied correctly to both numerator and denominator.
• Understanding Continuity: Always remember that continuity at a point means LHL = RHL = Function Value.

Summary

The problem requires us to use the condition of continuity of a piecewise function at $x=0$. We calculated the Right-Hand Limit (RHL) and the Left-Hand Limit (LHL) of the function as $x$ approaches 0 from the right and left sides, respectively. By equating these limits to the function's value at $x=0$ (which is $k$), we derived relationships between $a$, $b$, and $k$. Specifically, we found that $\frac{1}{a} + \frac{1}{b} = -4$ and $k = -4$. Substituting these values into the expression $\frac{1}{a} + \frac{1}{b} + \frac{4}{k}$ yielded the final answer.

The final answer is $\boxed{-5}$.