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

Unmasking the $1^{\infty }$ Trap

Welcome, future engineer. Today, we are going to dissect a problem that looks deceptively simple but hides a profound mathematical truth.

We are looking at the limit:

At first glance, your intuition might scream, 'The base is 1, and 1 to any power is 1!' But in the realm of calculus, intuition without rigor is a dangerous path. Let us embark on a journey to uncover why this limit is far more interesting than it appears.

Identifying the Indeterminate Form

Before we rush into calculations, we must diagnose the problem. Let us examine the base: ${\lim }_{x\to \infty }\frac{x-3}{x+2}$.

As $x$ grows towards infinity, the constants $-3$ and $+2$ become insignificant, like pebbles in an ocean. The ratio effectively becomes $\frac{x}{x}$, which is $1$.

Now, look at the exponent: $x$. As $x\to \infty$, the exponent itself grows without bound. We have arrived at the classic $1^{\infty }$ indeterminate form. This is not a value; it is a signal that we need a more sophisticated approach.

The Magic of the Exponential Rule

In JEE Advanced, time is your most precious currency. While you could use logarithms and L'Hopital's rule, there is a much more elegant path.

We use the standard result: if ${\lim }_{x\to a}f(x)=1$ and ${\lim }_{x\to a}g(x)=\infty$, then:

Think of this as a transformation. We are taking the 'exponential' nature of the problem and converting it into a 'multiplicative' one, all while keeping the base $e$ as our anchor. It is a beautiful bridge between algebra and calculus.

Algebraic Surgery

Now, let us apply this to our specific functions. Here, $f(x)=\frac{x-3}{x+2}$ and $g(x)=x$.

Our limit becomes:

The real work happens inside the bracket. We need to perform a bit of algebraic surgery by subtracting $1$ from our fraction:

Be careful here—the negative sign distributes to both terms in the bracket! This simplifies to:

The Final Convergence

We are almost there. We bring back the $x$ that was waiting outside the bracket. Our exponent limit is now:

To solve this, we divide the numerator and the denominator by $x$, the highest power present. We get:

As $x$ approaches infinity, the term $\frac{2}{x}$ vanishes into zero. We are left with $\frac{-5}{1}$, which is simply $-5$.

We have successfully tamed the exponent! The limit of the exponent is $-5$.

Now, we return to our base $e$. The final answer is $e^{-5}$.

Geometrically, this means that as $x$ moves toward infinity, the function ${\left(\frac{x-3}{x+2}\right)}^x$ settles down to the horizontal asymptote $y=e^{-5}$. It is a stunning result—a complex, shifting expression that eventually finds peace at a specific, transcendental value.