Animated Solution for Mathematics - Limits, Continuity and Differentiability: ${\lim }_{x\to \infty }{\left(\frac{x^2+5x+3}{x^2+x+3}\right)}^x$

Mastering the $1^{\infty }$ Limit

Welcome, fellow traveler of the mathematical landscape. Today, we stand before a classic sentinel of the JEE Advanced exam: the indeterminate form ${\lim }_{x\to \infty }{\left(\frac{x^2+5x+3}{x^2+x+3}\right)}^x$.

At first glance, this expression might seem intimidating. It is a function raised to the power of another function, and as $x$ grows without bound, the base approaches $1$ while the exponent explodes toward infinity.

This is the legendary $1^{\infty }$ form—a trap for the unwary, but a playground for the prepared.

The Diagnostic

Before we dive into the algebra, we must diagnose the patient. Let us examine the base: $f(x)=\frac{x^2+5x+3}{x^2+x+3}$.

As $x$ approaches infinity, we divide the numerator and denominator by the highest power, $x^2$. We see that:

As $x\to \infty$, all terms with $x$ in the denominator vanish, leaving us with $1/1=1$. Meanwhile, the exponent $g(x)=x$ is clearly racing toward infinity. We have confirmed our target: $1^{\infty }$.

The Elegant Shortcut

Many students attempt to use L'Hopital's rule immediately, leading to a forest of derivatives that often obscures the path to the solution. Instead, we invoke the elegant identity: if ${\lim }_{x\to a}f(x)=1$ and ${\lim }_{x\to a}g(x)=\infty$, then:

This formula is our compass. It transforms a daunting power-based limit into a simple product of a rational function and a polynomial.

The Algebraic Dance

Now, let us perform the subtraction $f(x)-1$. This is where the magic happens. We write:

By finding a common denominator, we get:

Watch closely as the $x^2$ terms and the constant $3$ terms cancel out with surgical precision. We are left with:

This is the heart of the problem. The complexity has been stripped away, leaving a clean, manageable fraction.

The Final Convergence

We now multiply this result by our exponent $g(x)=x$ to find $L$:

To evaluate this limit, we once again divide the numerator and denominator by $x^2$:

As $x$ marches toward infinity, the terms $1/x$ and $3/x^2$ dissolve into zero. We are left with $L=4/1=4$.

The final step is to return to our identity: the original limit is $e^L$. Substituting our value of $L$, we arrive at the beautiful, concise result: $e^4$.

Do you see the beauty here? We didn't fight the complexity; we transformed it.

By recognizing the $1^{\infty }$ structure, we bypassed the chaos and arrived at a result that feels almost inevitable. Keep this logic in your toolkit—whenever you see a function approaching $1$ raised to an infinite power, remember the $e^L$ transformation.

You have the power to simplify the universe, one limit at a time.