Animated Solution for Mathematics - Limits, Continuity and Differentiability: If the function $f(x)=\{\begin{array}{ll}\frac{{\log }_e(1-x+x^2)+{\log }_e(1+x+x^2)}{\sec x-\cos x},& x\in (-\frac{\pi }{2},\frac{\pi }{2})-\{0\}\\ k,& x=0\end{array}$ is continuous at $x=0$, then $k$ is equal to :

To ensure the function $f(x)$ is continuous at $x=0$, the function must satisfy the condition $k={\lim }_{x\to 0}f(x)$. This ensures there is no "hole" in the graph, as the limit from both sides matches the defined value of the function.

Our objective is to evaluate the limit of the given expression as $x$ approaches $0$.

Consider the numerator: ${\log }_e(1-x+x^2)+{\log }_e(1+x+x^2)$. Using the logarithmic property ${\log }_{e}A+{\log }_{e}B={\log }_e(AB)$, we combine the terms:

By rearranging the terms inside the brackets as $((1+x^2)-x)((1+x^2)+x)$, we apply the difference of squares identity $(A-B)(A+B)=A^2-B^2$, where $A=1+x^2$ and $B=x$.

Expanding this, we obtain:

Thus, the numerator simplifies to ${\log }_e(1+x^2+x^4)$.

Next, we address the denominator: $\sec x-\cos x$. Converting to the language of sines and cosines, we have:

Using the fundamental identity $1-{\cos }^{2}x={\sin }^{2}x$, the denominator becomes:

We now substitute these simplified forms into our limit expression:

To evaluate this, we utilize the standard limits ${\lim }_{u\to 0}\frac{{\log }_e(1+u)}{u}=1$ and ${\lim }_{x\to 0}\frac{\sin x}{x}=1$. We rewrite the expression to isolate these forms:

Evaluating each component as $x\to 0$:

1. ${\lim }_{x\to 0}\frac{{\log }_e(1+x^2+x^4)}{x^2+x^4}=1$ 2. ${\lim }_{x\to 0}\frac{x^2+x^4}{x^2}={\lim }_{x\to 0}(1+x^2)=1$ 3. ${\lim }_{x\to 0}\frac{x^2}{{\sin }^{2}x}=1$ 4. ${\lim }_{x\to 0}\cos x=1$

Multiplying these results, we find the final value: