Animated Solution for Mathematics - Limits, Continuity and Differentiability: ${\lim }_{x\to 2}\frac{3^x+3^{3-x}-12}{3^{-x/2}-3^{1-x}}$ is equal to.....

Welcome, student. Today, we are going to dismantle a problem that, at first glance, looks like a chaotic mess of exponential functions. You see terms like $3^x$ and $3^{3-x}$, and your instinct might be to reach for L'Hopital's Rule.

But hold on. In the world of JEE Advanced, we don't just solve problems; we master them. We look for the underlying structure. Let's embark on this journey to solve:

The Diagnostic

Before we touch any algebra, we must diagnose the patient. What is the nature of this limit? We substitute $x=2$ directly into the expression.

The numerator becomes $3^2+3^{3-2}-12=9+3-12=0$.

The denominator becomes $3^{-2/2}-3^{1-2}=3^{-1}-3^{-1}=0$.

We have arrived at the classic $\frac{0}{0}$ indeterminate form. This is not a wall; it is a doorway. It tells us that there is a 'hidden' factor lurking in both the numerator and the denominator—a factor that is causing the zero. Our mission is to find it and eliminate it.

The Strategic Substitution

When you see multiple powers of the same base, especially fractional ones, substitution is your best friend. Look at the exponents: $x$, $3-x$, $-x/2$, and $1-x$. They are all related to $x/2$.

Let us define a new variable, $t$, such that $t=3^{x/2}$.

This is a powerful move. It transforms our exponential landscape into a polynomial one. If $t=3^{x/2}$, then squaring both sides gives us $3^x=t^2$. Similarly, $3^{-x}$ becomes $\frac{1}{t^2}$.

But wait! We must never forget the boundary. If $x\to 2$, then $t=3^{2/2}=3^1=3$. So, our new limit is ${\lim }_{t\to 3}$.

If you forget to change the limit, you are solving for the wrong point. Always keep your boundary in sync with your variable.

The Algebraic Cleanup

Now, let's rewrite our expression in terms of $t$.

The numerator was $3^x+3^{3-x}-12$. Using our substitution, this becomes:

The denominator was $3^{-x/2}-3^{1-x}$. This becomes:

We now have a complex fraction:

This looks messy, but we have a secret weapon: multiply the numerator and the denominator by $t^2$. This will clear the fractions instantly.

Multiplying the numerator by $t^2$ gives us $t^4-12t^2+27$.

Multiplying the denominator by $t^2$ gives us $t-3$.

The Final Factorization

We are left with:

Look at that numerator. It is a bi-quadratic, but it behaves exactly like a quadratic if you treat $t^2$ as a single variable. We need two numbers that multiply to $27$ and add to $-12$. Those numbers are $-9$ and $-3$.

So, the numerator factors into $(t^2-9)(t^2-3)$. And we know that $(t^2-9)$ is a difference of squares: $(t-3)(t+3)$.

Our expression is now:

The Victory

Do you see it? The $(t-3)$ term in the numerator and denominator—the very culprit that caused the $0/0$ form—cancels out perfectly. We are left with ${\lim }_{t\to 3}(t+3)(t^2-3)$.

Now, we simply substitute $t=3$:

And there it is. The complexity vanishes, leaving behind a clean, elegant integer. Remember this process: diagnose, substitute, simplify, and factor. You have the tools to conquer any limit. The final answer is 36.