lim x→1 (∫₀^((x-1)²) t cos (t²) dt) / ((x-1) sin(x-1))

Q: Wait, why do we substitute $y = x-1$? Can't we just use L'Hospital's rule directly on $x$?

You absolutely could, but it would be a nightmare! Differentiating $(x-1) \sin(x-1)$ using the product rule is fine, but differentiating the integral with respect to $x$ when the limit is $(x-1)^2$ is prone to chain rule errors. Substituting $y = x-1$ simplifies the expression significantly, making the calculus much cleaner and less prone to silly mistakes.

Q: How do I know if I should use L'Hospital's rule or just simplify?

That is a great question! Always check the form first. If you get $\frac{0}{0}$ or $\frac{\infty}{\infty}$, L'Hospital is a valid tool. However, if you see terms like $\sin y$ or $\tan y$ in the denominator, it is often faster to use standard limits (like $\frac{\sin y}{y} \to 1$) rather than differentiating again. In this problem, dividing by $y$ turns the denominator into a form where the standard limit is obvious.

Q: Why does the lower limit of the integral disappear when I differentiate?

Remember the Newton-Leibniz rule: $\frac{d}{dy} \int_{g(y)}^{h(y)} f(t) dt = f(h(y)) \cdot h'(y) - f(g(y)) \cdot g'(y)$. In our case, the lower limit $g(y)$ is $0$, which is a constant. The derivative of a constant is $0$, so the term $f(g(y)) \cdot g'(y)$ becomes $f(0) \cdot 0$, which is just $0$. That is why it vanishes!

Question

$\lim_{x 	o 1} \frac{\int_0^{(x-1)^2} t \cos (t^2) dt}{(x-1) \sin(x-1)}$

Sigma Percentile · Accepted Answer

### The Art of Simplifying the Intimidating

Have you ever looked at a math problem and felt like you were staring at a locked door? The expression

$$\lim_{x 	o 1} \frac{\int_0^{(x-1)^2} t \cos (t^2) dt}{(x-1) \sin(x-1)}$$

is exactly that kind of door. It looks heavy, complex, and frankly, a bit scary.

But in JEE Advanced, the most intimidating problems often have the most elegant keys. Today, we are going to find that key together.

### Phase 1: The Power of Substitution

When you see the same expression repeating itself—in this case, $(x-1)$—your brain should immediately light up. It is a signal that the complexity is artificial.

Let us perform a substitution: let $y = x-1$. As $x 	o 1$, $y 	o 0$.

Suddenly, our limit becomes:

$$\lim_{y 	o 0} \frac{\int_0^{y^2} t \cos (t^2) dt}{y \sin y}$$

Look at that! The fog has cleared. We are no longer dealing with a complex $x$ expression; we are dealing with a clean, manageable $y$ expression.

### Phase 2: The Newton-Leibniz Tool

Now, we check the form. Plugging in $y=0$, the numerator becomes $\int_0^0 \dots = 0$, and the denominator becomes $0 \cdot \sin(0) = 0$.

We have a $\frac{0}{0}$ indeterminate form. It is time for L'Hospital's Rule.

To differentiate the integral in the numerator, we use the **Newton-Leibniz rule**:

$$\frac{d}{dy} \int_{0}^{h(y)} f(t) dt = f(h(y)) \cdot h'(y)$$

Here, $f(t) = t \cos(t^2)$ and $h(y) = y^2$. Substituting these in, the derivative of our numerator is:

$$f(y^2) \cdot \frac{d}{dy}(y^2) = (y^2) \cos((y^2)^2) \cdot 2y = 2y^3 \cos(y^4)$$

### Phase 3: The Elegant Finish

Now for the denominator: $\frac{d}{dy} (y \sin y)$. Using the product rule, we get $\sin y + y \cos y$.

Our limit is now:

$$\lim_{y 	o 0} \frac{2y^3 \cos(y^4)}{\sin y + y \cos y}$$

If we tried L'Hospital again, we would be stuck in a cycle of differentiation. Instead, let us be clever and divide both the numerator and the denominator by $y$:

$$\lim_{y 	o 0} \frac{2y^2 \cos(y^4)}{\frac{\sin y}{y} + \cos y}$$

As $y 	o 0$, we know that $\frac{\sin y}{y} 	o 1$ and $\cos(0) 	o 1$. The denominator becomes $1+1=2$.

The numerator becomes $2(0)^2 \cdot 1 = 0$. Thus, the final result is:

$$\frac{0}{2} = \mathbf{0}$$

We have successfully unlocked the door. Remember, in JEE, it is not about brute force; it is about finding the right tool and using it with grace.

Animated Solution for Mathematics - Definite Integration: $lim_{x \to 1} \frac{\int _{0}^{(x - 1)^{2}} t c o s ( t ^{2} ) d t}{( x - 1 ) s i n ( x - 1 )}$

Enter Numerical Value:

Visualized Solution

Analyze the Limit Expression

Apply Substitution $y = x - 1$

Rewrite the Limit in terms of $y$

Check the Limit Form

The Newton-Leibniz Tool

Differentiate the Numerator

Differentiate the Denominator

Re-assemble the Limit

Simplify using Standard Limits

Final Evaluation

The Sigma Insight: Newton-Leibniz & Reduction Formulas

The Art of Simplifying the Intimidating

Phase 1

Phase 2

Phase 3

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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→1​(x−1)sin(x−1)∫0(x−1)2​tcos(t2)dt​