If lim (x→0) (((a-n)nx - tan x) sin nx)/x² = 0, where n is nonzero real number, then a is equal to (

Q: Wait, why can't we just use L'Hopital's Rule here? It seems faster.

L'Hopital's Rule is a powerful tool, but it can be a trap in complex expressions like this. Differentiating the product of trigonometric functions often leads to messy, nested derivatives that are prone to calculation errors. Algebraic manipulation, as shown here, is cleaner, more elegant, and less likely to lead to mistakes.

Q: Why do we reject the possibility that $n=0$?

Great catch! The problem explicitly states that $n$ is a non-zero real number. In mathematics, always pay close attention to the constraints given in the problem statement. If $n$ were zero, the entire expression would collapse, and the limit would not be well-defined in the way we are analyzing.

Q: How do I know when to split the denominator like that?

A good rule of thumb is to look for products in the numerator. If you have a limit of the form $\lim_{x \to 0} \frac{f(x) \cdot g(x)}{x^2}$, it is almost always a sign that you should split it into $\lim_{x \to 0} \frac{f(x)}{x} \cdot \lim_{x \to 0} \frac{g(x)}{x}$. This 'divide and conquer' strategy simplifies the problem significantly.

Question

If $\lim_{x 	o 0} \frac{((a-n)nx - 	an x) \sin nx}{x^2} = 0$, where $n$ is nonzero real number, then $a$ is equal to

Options:

(A) 0

(B) $(n+1)/n$

(C) $n$

(D) $n + 1/n$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are standing before this limit:

$$\lim_{x 	o 0} \frac{((a-n)nx - 	an x) \sin nx}{x^2} = 0$$

At first glance, it looks intimidating. A complex numerator, a denominator of $x^2$, and a variable $a$ that we need to find. But in the world of JEE Advanced, complexity is often just a mask for a beautiful, simple structure waiting to be revealed.

### The Art of Splitting

The first thing that should catch your eye is the denominator, $x^2$. In calculus, when you see a denominator that can be factored, it is often a hint to distribute it.

We have two distinct parts in the numerator: the bracketed term $((a-n)nx - 	an x)$ and the term $\sin nx$. By splitting the $x^2$ into $x \cdot x$, we can rewrite the limit as a product of two simpler limits:

$$\lim_{x 	o 0} \left[ \frac{(a-n)nx - 	an x}{x} \cdot \frac{\sin nx}{x} ight] = 0$$

This is our "divide and conquer" moment. By separating the expression, we have transformed one terrifying problem into two manageable ones.

### The Trigonometric Toolkit

Now, let's focus on the second part: $\lim_{x 	o 0} \frac{\sin nx}{x}$. We know the standard limit $\lim_{	heta 	o 0} \frac{\sin 	heta}{	heta} = 1$.

Our expression is almost there, but the angle is $nx$, not $x$. The fix is simple: multiply and divide by $n$. This gives us:

$$\lim_{x 	o 0} \left( n \cdot \frac{\sin nx}{nx} ight)$$

As $x$ approaches zero, $nx$ also approaches zero, so $\frac{\sin nx}{nx}$ becomes $1$. Thus, the entire second part simplifies to $n \cdot 1 = n$.

Now, let's look at the first part: $\lim_{x 	o 0} \frac{(a-n)nx - 	an x}{x}$. We can split this into two fractions:

$$\lim_{x 	o 0} \left( \frac{(a-n)nx}{x} - \frac{	an x}{x} ight)$$

In the first term, the $x$ cancels out, leaving us with $(a-n)n$. In the second term, we use the standard limit $\lim_{x 	o 0} \frac{	an x}{x} = 1$. So, the first part simplifies to $(a-n)n - 1$.

### The Algebraic Resolution

We have successfully navigated the calculus. Now, we are in the realm of pure algebra. Our original limit equation has become:

$$((a-n)n - 1) \cdot n = 0$$

Since we know $n 
eq 0$, we can safely divide both sides by $n$, leaving us with $(a-n)n - 1 = 0$. From here, it is just a matter of isolating $a$.

We add $1$ to both sides to get $(a-n)n = 1$, then divide by $n$ to get $a - n = \frac{1}{n}$. Finally, adding $n$ to both sides gives us our final answer:

**$$a = n + \frac{1}{n}$$**

It is a beautiful result, isn't it? What started as a daunting limit problem resolved into a clean, elegant expression. Remember, in mathematics, the most complex problems often yield to the simplest, most fundamental principles.

Animated Solution for Mathematics - Limits, Continuity and Differentiability: If $lim_{x \to 0} \frac{(( a - n ) n x - t a n x ) s i n n x}{x ^{2}} = 0$ , where $n$ is nonzero real number, then $a$ is equal to

Select Answer:

Visualized Solution

Given Limit Expression

Standard Trigonometric Limits

Splitting the Denominator

Evaluating $lim_{x \to 0} \frac{s i n n x}{x}$

Splitting the First Term

Evaluating the First Term

Forming the Algebraic Equation

Applying the Non-Zero Condition

Transposing the Constant

Isolating the Bracket

Finding the Value of $a$

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

Analyzing the Setup

The Art of Splitting

The Trigonometric Toolkit

The Algebraic Resolution

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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; } If limx→0​x2((a−n)nx−tanx)sinnx​=0, where n is nonzero real number, then a is equal to