Animated Solution for Mathematics - Limits, Continuity and Differentiability: If the function $f(x)=\{\begin{array}{ll}\frac{1}{x}{\log }_e\left(\frac{1+\frac{x}{a}}{1-\frac{x}{b}}\right),& x<0\\ k,& x=0\\ \frac{{\cos }^{2}x-{\sin }^{2}x-1}{\sqrt{x^2+1}-1},& x>0\end{array}$ is continuous at $x=0$, then $\frac{1}{a}+\frac{1}{b}+\frac{4}{k}$ is equal to :

A Mathematical Journey

Imagine you are walking along a path defined by a function $f(x)$. For this path to be continuous at $x=0$, there must be no sudden jumps or gaps.

You should be able to walk from the left side, pass through the point at $x=0$, and continue to the right side without ever lifting your feet. This is the essence of continuity: the left-hand limit, the right-hand limit, and the value of the function at the point must all converge to the same destination, $k$.

Let us embark on this journey to find the value of $\frac{1}{a}+\frac{1}{b}+\frac{4}{k}$.

Conquering the Right Hand Limit

We begin our journey on the right side of the origin, where $x>0$. The function is defined as:

We recognize the numerator as a variation of the double angle identity. We know that ${\cos }^{2}x-{\sin }^{2}x=\cos 2x$. Thus, our numerator simplifies to $\cos 2x-1$.

Recalling our trigonometric toolkit, we know that $1-\cos 2x=2{\sin }^{2}x$. Therefore, $\cos 2x-1=-2{\sin }^{2}x$. Now, our expression is much cleaner:

To handle the denominator, we use the classic technique of rationalization. We multiply the numerator and denominator by the conjugate, $\sqrt{x^2+1}+1$. This transforms the denominator into $(\sqrt{x^2+1}{)}^2-1^2=x^2+1-1=x^2$.

Now, our limit looks like this:

By grouping $\frac{{\sin }^{2}x}{x^2}$ as $(\frac{\sin x}{x}{)}^2$, we can apply the standard limit ${\lim }_{x\to 0}\frac{\sin x}{x}=1$. Substituting $x=0$ into the remaining part, we get:

The right-hand limit is $-4$, which means our function must also be $-4$ at $x=0$. Thus, $k=-4$.

Decoding the Left Hand Limit

Now, we turn to the left side, where $x<0$. The function is:

Using the property $\log (M/N)=\log M-\log N$, we split this into:

To solve this, we use the standard limit ${\lim }_{u\to 0}\frac{{\log }_e(1+u)}{u}=1$. For the first term, we multiply and divide by $a$ to get $\frac{1}{a}\cdot \frac{{\log }_e(1+x/a)}{x/a}$, which approaches $\frac{1}{a}\times 1$.

For the second term, we multiply and divide by $-b$ to get $\frac{1}{b}\cdot \frac{{\log }_e(1-x/b)}{-x/b}$, which approaches $\frac{1}{b}\times 1$. Adding these together, the left-hand limit is $\frac{1}{a}+\frac{1}{b}$.

The Grand Synthesis

We have reached the final stage of our journey. We know that for continuity, the left-hand limit must equal $k$.

Since $k=-4$, we have $\frac{1}{a}+\frac{1}{b}=-4$. The question asks us to evaluate $\frac{1}{a}+\frac{1}{b}+\frac{4}{k}$.

Substituting our known values, we get:

We have successfully navigated the path and arrived at our destination. The final result is $-5$.