Key Concepts and Formulas

• Continuity of a Function: A function $f(x)$ is continuous at a point $x=a$ if $\mathop {\lim }\limits_{x \to a} f(x) = f(a)$.
• Limit of Logarithmic Functions: The standard limit $\mathop {\lim }\limits_{y \to 0} \frac{\ln(1+y)}{y} = 1$.
• Algebraic Manipulation of Limits: Techniques like factorization, rationalization, and using standard limits are crucial for evaluating indeterminate forms.
• Trigonometric Identities: $\sec x = \frac{1}{\cos x}$, and $\sin^2 x + \cos^2 x = 1$.

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 a function to be continuous at a point $x=a$, the limit of the function as $x$ approaches $a$ must exist and be equal to the value of the function at $a$. In this case, $a=0$, so we must have: $\mathop {\lim }\limits_{x \to 0} f(x) = f(0)$ From the definition of $f(x)$, we know that $f(0) = k$. Therefore, we need to find the limit of $f(x)$ as $x \to 0$ and set it equal to $k$.

Step 2: Set up the Limit Expression We need to evaluate the limit of $f(x)$ as $x \to 0$ using the definition for $x \neq 0$: $k = \mathop {\lim }\limits_{x \to 0} \frac{{{{\log }_e}(1 - x + {x^2}) + {{\log }_e}(1 + x + {x^2})}}{{\sec x - \cos x}}$

Step 3: Simplify the Numerator using Logarithm Properties Using the property $\log a + \log b = \log(ab)$, we can combine the logarithmic terms in the numerator: $(1 - x + x^2)(1 + x + x^2)$ This expression can be rearranged as $((x^2+1) - x)((x^2+1) + x)$. This is in the form $(a-b)(a+b) = a^2 - b^2$, where $a = x^2+1$ and $b = x$. $(x^2+1)^2 - x^2 = (x^4 + 2x^2 + 1) - x^2 = x^4 + x^2 + 1$ So, the numerator becomes ${{\log }_e}({x^4} + {x^2} + 1)$. The limit expression is now: $k = \mathop {\lim }\limits_{x \to 0} \frac{{{{\log }_e}({x^4} + {x^2} + 1)}}{{\sec x - \cos x}}$

Step 4: Simplify the Denominator using Trigonometric Identities First, express $\sec x$ in terms of $\cos x$: $\sec x = \frac{1}{\cos x}$. The denominator becomes: $\frac{1}{\cos x} - \cos x = \frac{1 - \cos^2 x}{\cos x}$ Using the identity $\sin^2 x + \cos^2 x = 1$, we have $1 - \cos^2 x = \sin^2 x$. So, the denominator simplifies to $\frac{\sin^2 x}{\cos x}$.

Step 5: Substitute the Simplified Denominator back into the Limit Now, substitute the simplified denominator into the limit expression: $k = \mathop {\lim }\limits_{x \to 0} \frac{{{{\log }_e}({x^4} + {x^2} + 1)}}{{\frac{\sin^2 x}{\cos x}}}$ $k = \mathop {\lim }\limits_{x \to 0} \frac{{\cos x \cdot {{\log }_e}({x^4} + {x^2} + 1)}}{{\sin^2 x}}$

Step 6: Identify the Indeterminate Form As $x \to 0$:

• $\cos x \to \cos 0 = 1$.
• ${{\log }_e}({x^4} + {x^2} + 1) \to {{\log }_e}(0^4 + 0^2 + 1) = {{\log }_e}(1) = 0$.
• $\sin^2 x \to \sin^2 0 = 0$. The limit is of the indeterminate form $\frac{1 \cdot 0}{0}$, which is essentially $\frac{0}{0}$. This indicates we need further manipulation.

Step 7: Rearrange the Limit for Standard Forms We want to use the standard limit $\mathop {\lim }\limits_{y \to 0} \frac{\ln(1+y)}{y} = 1$. Let's rewrite the expression: $k = \mathop {\lim }\limits_{x \to 0} \cos x \cdot \mathop {\lim }\limits_{x \to 0} \frac{{{{\log }_e}({x^4} + {x^2} + 1)}}{{\sin^2 x}}$ We know $\mathop {\lim }\limits_{x \to 0} \cos x = 1$. So, we focus on the second part. $k = 1 \cdot \mathop {\lim }\limits_{x \to 0} \frac{{{{\log }_e}({x^4} + {x^2} + 1)}}{{\sin^2 x}}$ To use the standard logarithmic limit, we need the argument of the logarithm to be in the form $1+y$, where $y \to 0$. Here, the argument is $x^4 + x^2 + 1$. We can write this as $1 + (x^4 + x^2)$. So, $y = x^4 + x^2$. We need to divide and multiply by $x^4 + x^2$ in the numerator: $\mathop {\lim }\limits_{x \to 0} \frac{{{{\log }_e}(1 + (x^4 + x^2))}}{{(x^4 + x^2)}} \cdot \frac{(x^4 + x^2)}{{\sin^2 x}}$ By the standard limit, $\mathop {\lim }\limits_{x \to 0} \frac{{{{\log }_e}(1 + (x^4 + x^2))}}{{(x^4 + x^2)}} = 1$, because as $x \to 0$, $x^4 + x^2 \to 0$.

Step 8: Evaluate the Remaining Part of the Limit Now we need to evaluate: $\mathop {\lim }\limits_{x \to 0} \frac{(x^4 + x^2)}{{\sin^2 x}}$ We can factor out $x^2$ from the numerator: $\mathop {\lim }\limits_{x \to 0} \frac{{x^2(x^2 + 1)}}{{\sin^2 x}}$ Rearrange this as: $\mathop {\lim }\limits_{x \to 0} \left(\frac{x^2}{{\sin^2 x}}\right) \cdot \mathop {\lim }\limits_{x \to 0} (x^2 + 1)$ We know the standard limit $\mathop {\lim }\limits_{x \to 0} \frac{\sin x}{x} = 1$. Therefore, $\mathop {\lim }\limits_{x \to 0} \frac{x}{\sin x} = 1$, and $\mathop {\lim }\limits_{x \to 0} \frac{x^2}{\sin^2 x} = \left(\mathop {\lim }\limits_{x \to 0} \frac{x}{\sin x}\right)^2 = 1^2 = 1$. The second part of the limit is: $\mathop {\lim }\limits_{x \to 0} (x^2 + 1) = 0^2 + 1 = 1$

Step 9: Combine the Results to Find k Putting all the pieces together: $k = 1 \cdot \left( \mathop {\lim }\limits_{x \to 0} \frac{{{{\log }_e}(1 + (x^4 + x^2))}}{{(x^4 + x^2)}} \right) \cdot \left( \mathop {\lim }\limits_{x \to 0} \frac{{x^2}}{{\sin^2 x}} \right) \cdot \left( \mathop {\lim }\limits_{x \to 0} (x^2 + 1) \right)$ $k = 1 \cdot 1 \cdot 1 \cdot 1$ $k = 1$

Common Mistakes & Tips

• Incorrectly applying standard limits: Ensure the argument of the logarithm is in the form $1+y$ and that $y \to 0$. Similarly, for $\frac{\sin x}{x}$, ensure the numerator and denominator match as $x \to 0$.
• Algebraic errors: Be careful when simplifying the numerator and denominator, especially with squares and products of terms. A small mistake here can lead to an incorrect final answer.
• Ignoring the $\cos x$ term: In Step 5, it's easy to forget the $\cos x$ term that arises from simplifying the denominator. This term evaluates to 1 as $x \to 0$, but it must be included in the calculation.

Summary

For the function $f(x)$ to be continuous at $x=0$, the limit of $f(x)$ as $x$ approaches $0$ must be equal to $f(0)$, which is $k$. We evaluated the limit by first simplifying the numerator using logarithm properties and the denominator using trigonometric identities. The resulting limit was of an indeterminate form, which we resolved by strategically multiplying and dividing by terms to utilize the standard limits $\mathop {\lim }\limits_{y \to 0} \frac{\ln(1+y)}{y} = 1$ and $\mathop {\lim }\limits_{x \to 0} \frac{\sin x}{x} = 1$. After careful algebraic manipulation and application of these standard limits, we found that $k$ must be equal to 1.

The final answer is \boxed{1}.