Animated Solution for Mathematics - Limits, Continuity and Differentiability: Let $f:(0,\pi )\to \mathbb{R}$ be a function given by $f(x)=\{\begin{array}{ll}(\frac{8}{7}{)}^{\frac{\tan 8x}{\tan 7x}},& 0<x<\frac{\pi }{2}\\ a-8,& x=\frac{\pi }{2}\\ (1+|\cot x|{)}^{\frac{b}{a}|\tan x|},& \frac{\pi }{2}<x<\pi \end{array}$ where $a,b\in \mathbb{Z}$. If $f$ is continuous at $x=\frac{\pi }{2}$, then $a^2+b^2$ is equal to

Imagine you are an architect of a bridge. In the world of calculus, continuity is the structural integrity of that bridge.

If a function is continuous at a point, it means there is no gap, no jump, and no hole. For our function $f(x)$ to be continuous at $x=\frac{\pi }{2}$, the path from the left, the path from the right, and the point itself must all converge to the exact same value.

It is a beautiful, harmonious meeting of three distinct mathematical journeys.

A Trigonometric Dance

Let us first walk from the left. As $x$ approaches $\frac{\pi }{2}$ from the left, we are dealing with the expression $(\frac{8}{7}{)}^{\frac{\tan 8x}{\tan 7x}}$.

Let us simplify it with a substitution. Let $x=\frac{\pi }{2}-h$, where $h$ is a tiny positive value approaching zero. As $x\to {\frac{\pi }{2}}^{-}$, $h\to 0$.

Now, look at the exponent:

Using our trigonometric identities, the numerator becomes $\tan (4\pi -8h)=-\tan 8h$. The denominator becomes $\tan (\frac{7\pi }{2}-7h)=\cot 7h$.

Since $\cot 7h=\frac{1}{\tan 7h}$, our exponent transforms into $-\tan 8h\cdot \tan 7h$. As $h\to 0$, both $\tan 8h$ and $\tan 7h$ vanish to zero.

Thus, the entire exponent becomes zero. Any non-zero base raised to the power of zero is $1$. Our left-hand limit is $1$. The bridge is holding steady.

Taming the Indeterminate

Now, let us walk from the right. As $x$ approaches $\frac{\pi }{2}$ from the right, we face $(1+|\cot x|{)}^{\frac{b}{a}|\tan x|}$.

As $x\to {\frac{\pi }{2}}^{+}$, $|\cot x|\to 0$ and $|\tan x|\to \infty$. This is the classic $1^{\infty }$ indeterminate form.

We have a powerful tool:

Applying this, our limit becomes:

Here is the magic: $|\cot x|\cdot |\tan x|=1$. The entire expression simplifies to $e^{\frac{b}{a}}$. The complexity vanishes, leaving us with a clean, elegant result.

We have our three values: the left-hand limit is $1$, the function value at $x=\frac{\pi }{2}$ is $a-8$, and the right-hand limit is $e^{\frac{b}{a}}$. For continuity, they must all be equal:

Solving $a-8=1$ gives us $a=9$. Solving $e^{\frac{b}{a}}=1$ implies $\frac{b}{a}=0$, which means $b=0$.

We have found our integers! The final step is to calculate:

We have successfully built our bridge, ensuring every piece fits perfectly. Calculus is not just about solving equations; it is about finding the hidden order in the chaos.