Animated Solution for Mathematics - Limits, Continuity and Differentiability: Let $f:(-\frac{\pi }{4},\frac{\pi }{4})\to \mathbb{R}$ be defined as $f(x)=\{\begin{array}{ll}(1+|\sin x|{)}^{\frac{3a}{|\sin x|}},& -\frac{\pi }{4}<x<0\\ b,& x=0\\ e^{\cot 4x/\cot 2x},& 0<x<\frac{\pi }{4}\end{array}$ If $f$ is continuous at $x=0$, then the value of $6a+b^2$ is equal to:

A Journey into Limits

Imagine you are an engineer tasked with building a bridge across a canyon. The canyon is the point $x=0$. On the left side, you have one construction team building from the negative side; on the right, another team building from the positive side.

For the bridge to be safe—for the function to be continuous—these two teams must meet at the exact same height, $b$, at the center. If they don't, the bridge collapses. This is the essence of continuity:

Let us walk through this construction together.

The Left Hand Limit (LHL) - The Modulus Trap

As we approach $x=0$ from the left, we encounter the function $f(x)=(1+|\sin x|{)}^{\frac{3a}{|\sin x|}}$. Here is where many students stumble. The absolute value $|\sin x|$ is not just a symbol; it is a gatekeeper.

Since $x$ is approaching $0$ from the negative side, $x$ is in the fourth quadrant where $\sin x$ is negative. Thus, $|\sin x|=-\sin x$.

Substituting this, our LHL becomes ${\lim }_{x\to 0^{-}}(1-\sin x{)}^{\frac{3a}{-\sin x}}$. To make this look like our standard limit, let us perform a substitution. Let $u=-\sin x$. As $x\to 0^{-}$, $u$ approaches $0^{+}$.

Now, the expression transforms into ${\lim }_{u\to 0^{+}}(1+u{)}^{\frac{3a}{u}}$. Does this look familiar? It is the classic form ${\lim }_{u\to 0}(1+u{)}^{\frac{1}{u}}=e$. Raising this to the power of $3a$, we get our LHL: $e^{3a}$. We have successfully bridged the left side!

The Right Hand Limit (RHL) - The Trigonometric Transformation

Now, we turn to the right side. The function is $f(x)=e^{\frac{\cot 4x}{\cot 2x}}$. Cotangents can be messy, so let us simplify. Using the identity $\cot \theta =\frac{1}{\tan \theta }$, the exponent becomes $\frac{\tan 2x}{\tan 4x}$.

We need to find the limit of this exponent as $x\to 0^{+}$. We use the standard limit ${\lim }_{\theta \to 0}\frac{\tan \theta }{\theta }=1$.

To force our expression into this form, we multiply and divide by the respective angles:

As $x\to 0$, the $\frac{\tan \theta }{\theta }$ terms vanish into $1$, leaving us with $\frac{2x}{4x}=\frac{1}{2}$. Thus, our RHL is $e^{1/2}$.

The Synthesis - Solving for the Unknowns

We have our pieces: $\text{LHL}=e^{3a}$, $\text{RHL}=e^{1/2}$, and $f(0)=b$. For continuity, these must be equal:

This gives us two beautiful equations. First, equating the exponents of the LHL and RHL: $3a=\frac{1}{2}$, which implies $6a=1$. Second, equating the RHL to $b$: $b=e^{1/2}$. Squaring both sides gives $b^2=e$.

Finally, the question asks for $6a+b^2$. Substituting our findings, we get $1+e$. The bridge is complete, the math is sound, and the solution is elegant.