Animated Solution for Mathematics - Limits, Continuity and Differentiability: ${\lim }_{x\to -\infty }\left[\frac{x^4\sin (1/x)+x^2}{(1+|x{|}^3)}\right]=.........$

A Limit Masterclass

Welcome, future engineer. Today, we are going to dissect a problem that, at first glance, looks like a chaotic mess of symbols. We have a rational function, a trigonometric term, and the dreaded modulus function.

But here is the secret to JEE Advanced: complexity is often just a mask for simplicity. Let us peel back the layers together.

The Modulus Trap

Our journey begins with the limit:

When you see $x\to -\infty$, your first instinct should be to define the behavior of your variables. We are not just dealing with any number; we are dealing with numbers that are negative and growing in magnitude without bound.

This is the moment where most students stumble. They see $|x|$ and think, 'Oh, it's just $x$.' But wait! If $x$ is negative, $|x|$ is not $x$. By definition, $|x|=-x$ when $x<0$.

So, our denominator, $1+|x{|}^3$, transforms. Since $|x|=-x$, then $|x{|}^3=(-x{)}^3=-x^3$. Suddenly, the denominator becomes $1-x^3$. The absolute value has vanished, replaced by a clean polynomial. This is the first victory in our battle.

The Algebraic Cleanup

Now, we have the expression:

When we face limits at infinity, our standard weapon of choice is to factor out the highest power of $x$. Why? Because it allows us to see which terms dominate and which terms fade away to zero.

Looking at the denominator, $1-x^3$, the highest power is clearly $x^3$. Let us be bold and factor $x^3$ out of both the numerator and the denominator.

In the numerator, we have $x^4\sin (1/x)+x^2$. If we pull out $x^3$, we are left with:

In the denominator, we have $1-x^3$. Factoring out $x^3$ gives us $x^3\left(\frac{1}{x^3}-1\right)$.

Now, look at the expression:

The $x^3$ terms cancel out perfectly! We are left with ${\lim }_{x\to -\infty }\frac{x\sin (1/x)+1/x}{1/x^3-1}$. The chaos is receding.

The Calculus Magic

We are almost there. We have three distinct parts to evaluate: $x\sin (1/x)$, $1/x$, and $1/x^3$.

Let us tackle the most interesting one: $x\sin (1/x)$. This is a classic JEE setup. If we let $\theta =1/x$, then as $x\to -\infty$, $\theta \to 0$.

The expression $x\sin (1/x)$ becomes $\frac{\sin (\theta )}{\theta }$. We know from our fundamental calculus toolkit that:

Next, consider the simpler terms. As $x$ grows to negative infinity, $1/x$ approaches $0$, and $1/x^3$ also approaches $0$.

Substituting these values back into our expression, we get:

And there it is. The limit is $-1$.

Notice how we didn't need complex rules or brute-force differentiation. We used the definition of the modulus, we used algebraic factoring to simplify the expression, and we used the standard limit of the sine function.

This is the beauty of mathematics—taking a seemingly impossible problem and breaking it down into small, manageable, and elegant steps. Keep this mindset, and you will conquer any problem the JEE throws at you.