lim (x→0) 48/x⁴ ∫₀ˣ t³/(t⁶+1) dt is equal to ______.

Q: Wait, why can't I just integrate $\frac{t^3}{t^6+1}$ directly? It looks like a standard substitution problem.

You absolutely could! If you substitute $u = t^3$, then $du = 3t^2 dt$. However, that would lead to an inverse tangent function involving $x^3$. While that is mathematically valid, it makes the limit significantly harder to evaluate. In JEE, we look for the 'path of least resistance'—L'Hopital's Rule is designed specifically to bypass the need for explicit integration.

Q: Why does the derivative of the integral $\int_{0}^{x} \frac{t^3}{t^6+1} dt$ just become $\frac{x^3}{x^6+1}$?

That is the magic of the Leibniz Integral Rule! When you differentiate an integral with respect to its upper limit $x$, you are essentially asking for the rate at which the area under the curve grows as the boundary moves. The rule states that the derivative is simply the integrand evaluated at the upper limit multiplied by the derivative of that limit. Since the upper limit is just $x$, its derivative is $1$.

Q: Is it always safe to use L'Hopital's Rule whenever I see a fraction?

Great question! No, it is not. You must verify the indeterminate form ($\frac{0}{0}$ or $\frac{\infty}{\infty}$) first. If the limit is not indeterminate, applying L'Hopital's will lead you to a completely wrong answer. Always check the limit of the numerator and denominator separately before differentiating.

Q: What happens if I had $x^5$ in the denominator instead of $x^4$? Would the answer still be 12?

If the denominator were $x^5$, the $x^3$ terms would not cancel out completely. You would be left with an $x^2$ in the denominator, which would cause the limit to approach infinity as $x \to 0$. The power of $x$ in the denominator is perfectly tuned to the power of $t$ in the integrand to ensure a finite, non-zero limit.

Q: I always get confused with the constants. Do I include the 48 in the derivative?

Yes! The constant $48$ is a multiplier. When you differentiate $48 \cdot F(x)$, you get $48 \cdot F'(x)$. Don't let the constant intimidate you; just treat it as a scalar that 'tags along' until the very end of the calculation.

Question

$\lim_{x 	o 0} \frac{48}{x^4} \int_{0}^{x} \frac{t^3}{t^6+1} dt$ is equal to ______.

Sigma Percentile · Accepted Answer

### The Elegance of the Indeterminate Form

Welcome, fellow traveler on the path to JEE mastery. Today, we are going to dissect a problem that, at first glance, might look like a daunting wall of calculus. We are tasked with evaluating the limit:

$$ \lim_{x 	o 0} \frac{48}{x^4} \int_{0}^{x} \frac{t^3}{t^6+1} dt $$

Take a deep breath. I know that integral looks intimidating. It sits there, nested inside a fraction, waiting to confuse us. But in the world of competitive mathematics, complexity is often just a mask for a very simple, elegant truth. Let us peel back that mask together.

### Step 1: The Diagnostic Check

Before we charge into battle, we must perform a diagnostic. As $x 	o 0$, the denominator $x^4$ clearly heads toward $0$.

The numerator, represented by the integral $\int_{0}^{x} \frac{t^3}{t^6+1} dt$, represents the area under a curve from $0$ to $x$. As $x$ shrinks to $0$, the interval of integration vanishes, and the area becomes $0$.

We have arrived at the classic $\frac{0}{0}$ indeterminate form. This is our green light—our signal that **L'Hopital's Rule** is the key to unlocking this puzzle.

### Step 2: The Power of Leibniz

Now, we need to differentiate the numerator. We use the **Newton-Leibniz Rule**, which states that the derivative of an integral with respect to its upper limit is simply the integrand evaluated at that limit.

Mathematically, we have:

$$ \frac{d}{dx} \int_{0}^{x} \frac{t^3}{t^6+1} dt = \frac{x^3}{x^6+1} $$

It is almost poetic; the integral sign and the derivative sign essentially 'cancel' each other out. We multiply this by our constant $48$, and our numerator derivative becomes $\frac{48x^3}{x^6+1}$.

### Step 3: The Dance of Cancellation

Now, let us look at the denominator. The derivative of $x^4$ is a simple application of the power rule: $4x^3$.

Putting it all together, our limit transforms into:

$$ \lim_{x 	o 0} \frac{\frac{48x^3}{x^6+1}}{4x^3} $$

Look closely at this expression. The $x^3$ term is present in both the numerator and the denominator. By canceling the $x^3$ terms, the 'undefined' nature of the limit evaporates. We are left with:

$$ \lim_{x 	o 0} \frac{48}{4(x^6+1)} $$

### The Final Victory

We are in the home stretch. We simplify the constants: $\frac{48}{4} = 12$.

Now, we simply evaluate the limit as $x 	o 0$. Substituting $x=0$ into our remaining expression, we get:

$$ \frac{12}{0^6+1} = \frac{12}{1} = 12 $$

And there it is. The complexity has collapsed into a single, clean integer. The final answer is **12**.

Animated Solution for Mathematics - Definite Integration: $lim_{x \to 0} \frac{48}{x ^{4}} \int_{0}^{x} \frac{t ^{3}}{t ^{6} + 1} d t$ is equal to ______.

Enter Numerical Value:

Visualized Solution

Analyzing the Limit Expression

Checking the $\frac{0}{0}$ Form

Strategy: L'Hopital's Rule

The Newton-Leibniz Formula

Setting up the Derivatives

Differentiating the Numerator

Differentiating the Denominator

Simplifying the Expression

Canceling Terms

Reducing the Constants

Evaluating the Final Limit

The Sigma Insight: Newton-Leibniz & Reduction Formulas

The Elegance of the Indeterminate Form

Step 1

Step 2

Step 3

The Final Victory

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Animated Solution for Mathematics - Definite Integration: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } limx→0​x448​∫0x​t6+1t3​dt is equal to ______.