f(x) is the integral of (2 sin x - sin 2x)/x³, x ≠ 0, find lim (x→0) f'(x)

Q: Wait, why can't I just integrate $\frac{2 \sin x - \sin 2x}{x^3}$ first and then differentiate?

That is a classic trap! The integral of this function cannot be expressed in terms of elementary functions. The Fundamental Theorem of Calculus is your shortcut—it tells you the derivative of an integral is just the integrand. Don't do the hard work when the theorem does it for you!

Q: I keep forgetting the limit of $\frac{1 - \cos x}{x^2}$. Is it $0$ or $\frac{1}{2}$?

It is $\frac{1}{2}$. Think of it this way: $1 - \cos x$ is approximately $\frac{x^2}{2}$ using the Taylor series expansion. So, $\frac{x^2/2}{x^2}$ gives you $\frac{1}{2}$. It's a crucial standard limit for JEE!

Q: Why do we need to split $x^3$ into $x$ and $x^2$? Can't we just use L'Hopital's Rule?

You absolutely can use L'Hopital's Rule! However, splitting the denominator into standard limits is often faster and less prone to calculation errors during differentiation. Both methods are valid, but mastering the standard limits makes you a faster problem solver.

Question

$f(x)$ is the integral of $\frac{2 \sin x - \sin 2x}{x^3}, x 
eq 0$, find $\lim_{x 	o 0} f'(x)$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

When you first look at the expression $f(x) = \int \frac{2 \sin x - \sin 2x}{x^3} dx$, your instinct might be to panic. However, the secret of the JEE Advanced is that the most intimidating problems often require the least amount of actual integration.

We are tasked with finding $\lim_{x 	o 0} f'(x)$.

### The Fundamental Theorem: Your Get-Out-of-Jail-Free Card

Many students fall into the trap of trying to find the antiderivative of the integrand. They reach for integration by parts or series expansion, wasting precious minutes.

The **Fundamental Theorem of Calculus** is our best friend here. It tells us that the derivative of an integral is simply the integrand itself:

$$\frac{d}{dx} \int g(x) dx = g(x)$$

By applying this, the integral sign vanishes, and we are left with the function:

$$f'(x) = \frac{2 \sin x - \sin 2x}{x^3}$$

Just like that, the complexity of the integral disappears, and we are left with a limit problem.

### The Trigonometric Dance

Now, we face the limit: $\lim_{x 	o 0} \frac{2 \sin x - \sin 2x}{x^3}$. If you plug in $x = 0$ immediately, you get the indeterminate form $\frac{0}{0}$.

We must unify the arguments in the numerator. Recall the double angle identity: $\sin 2x = 2 \sin x \cos x$.

Substituting this into our expression, the numerator transforms into:

$$2 \sin x - 2 \sin x \cos x = 2 \sin x (1 - \cos x)$$

The expression now becomes:

$$\lim_{x 	o 0} \frac{2 \sin x (1 - \cos x)}{x^3}$$

### The Surgical Precision of Limits

We have $x^3$ in the denominator and two distinct trigonometric parts in the numerator. We split $x^3$ into $x \cdot x^2$ to match the standard limits we know by heart.

We rewrite the limit as:

$$\lim_{x 	o 0} 2 \cdot \left( \frac{\sin x}{x} ight) \cdot \left( \frac{1 - \cos x}{x^2} ight)$$

We know that $\lim_{x 	o 0} \frac{\sin x}{x} = 1$ and $\lim_{x 	o 0} \frac{1 - \cos x}{x^2} = \frac{1}{2}$.

Substituting these values, we get:

$$2 \cdot 1 \cdot \frac{1}{2} = 1$$

### Final Reflection

We started with a terrifying integral, used the Fundamental Theorem to strip away the complexity, employed a trigonometric identity to reveal the structure, and used standard limits to find the truth.

The limit of the derivative as $x$ approaches zero is **$1$**.

Animated Solution for Mathematics - Limits, Continuity and Differentiability: $f (x)$ is the integral of $\frac{2 s i n x - s i n 2 x}{x ^{3}}, x \neq = 0$ , find $lim_{x \to 0} f^{'} (x)$

Enter Numerical Value:

Visualized Solution

The Problem Statement

Derivative of an Integral

Expression for $f^{'} (x)$

The Limit Problem

Double Angle Formula

Substituting the Identity

Factoring the Numerator

Rearranging Terms

Standard Limit Forms

Evaluating the Limit

Final Answer

The Sigma Insight: Evaluation of Limits & L'Hopital's Rule

Analyzing the Setup

The Fundamental Theorem

The Trigonometric Dance

The Surgical Precision of Limits

Final Reflection

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Animated Solution for Mathematics - Limits, Continuity and Differentiability: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } f(x) is the integral of x32sinx−sin2x​,x=0, find limx→0​f′(x)