Animated Solution for Mathematics - Limits, Continuity and Differentiability: ${\lim }_{x\to \infty }\frac{(2x^2-3x+5)(3x-1{)}^{\frac{x}{2}}}{(3x^2+5x+4)\sqrt{(3x+2{)}^x}}$ is equal to:

A Journey Through Complexity

Welcome, fellow traveler on the path to JEE excellence. Today, we are going to dismantle a problem that, at first glance, looks like a chaotic mess of variables and powers.

It is the kind of problem that makes your heart skip a beat—not because it is impossible, but because it is intimidating. But remember, in the world of mathematics, complexity is often just a disguise for simplicity. Let us peel back the layers of this limit together.

The Art of Decomposition

We are presented with the limit:

When you see a monster like this, your first instinct might be to panic. Don't. Instead, look for the structure.

We have a product of terms in the numerator and a product in the denominator. The square root in the denominator, $\sqrt{(3x+2{)}^x}$, is simply $(3x+2{)}^{\frac{x}{2}}$.

By grouping the algebraic terms and the exponential terms, we can rewrite the limit as a product of two smaller, much friendlier limits:

By splitting the problem into $L_1$ and $L_2$, we have already won half the battle. We have transformed one terrifying problem into two manageable tasks.

Taming the Algebraic Beast

Let us tackle $L_1={\lim }_{x\to \infty }\frac{2x^2-3x+5}{3x^2+5x+4}$. This is a classic rational function.

As $x$ grows to infinity, the terms with $x^2$ dominate the behavior of the function. To see this clearly, we divide every term in the numerator and denominator by $x^2$:

As $x$ approaches infinity, any term with $x$ in the denominator—like $\frac{3}{x}$ or $\frac{4}{x^2}$—vanishes into zero. We are left with the elegant ratio of the leading coefficients: $L_1=\frac{2}{3}$.

The $1^{\infty }$ Mystery

Now, we turn our attention to the exponential part: $L_2={\lim }_{x\to \infty }{\left(\frac{3x-1}{3x+2}\right)}^{\frac{x}{2}}$.

First, check the form. As $x\to \infty$, the base $\frac{3x-1}{3x+2}$ approaches $1$, and the exponent $\frac{x}{2}$ approaches $\infty$. This is the famous $1^{\infty }$ indeterminate form.

Whenever you see this, you must immediately reach for the standard identity:

Let us apply this. Here, $f(x)=\frac{3x-1}{3x+2}$ and $g(x)=\frac{x}{2}$. The exponent of our $e$ becomes:

Focus on the term inside the parenthesis. Finding a common denominator, we get $\frac{3x-1-(3x+2)}{3x+2}$, which simplifies beautifully to $\frac{-3}{3x+2}$.

Now, multiply this by the exponent $\frac{x}{2}$:

Again, we are looking at a limit at infinity. Divide by $x$, and we get $\frac{-3}{6}=-\frac{1}{2}$. Thus, $L_2=e^{-\frac{1}{2}}$, or $\frac{1}{\sqrt{e}}$.

The Grand Synthesis

We have our two pieces. $L_1=\frac{2}{3}$ and $L_2=\frac{1}{\sqrt{e}}$. Multiplying them together gives us the final answer:

Look at that. From a complex, intimidating expression, we have arrived at a concise, elegant result. This is the beauty of mathematics.

It rewards patience, structure, and the courage to break down the impossible into the possible. Keep practicing, keep questioning, and most importantly, keep falling in love with the process.