If f(x) = { (sin(a+2)x + sin x)/x, x 0 } is co

Q: Why can't we just plug in $x=0$ directly into the function to find the limit?

If you try to plug in $x=0$ directly, you will encounter the indeterminate form $0/0$. In calculus, the value of a function at a point and the limit of a function as it approaches that point are two different things. We use limits to understand the behavior of the function as it gets infinitely close to the point, bypassing the 'division by zero' trap.

Q: Is L'Hopital's rule better than binomial approximation for the right-hand limit?

L'Hopital's rule is a powerful weapon, and it would certainly work here. However, binomial approximation is often faster and more elegant for expressions involving radicals like $(1+3x)^{1/3}$. Mastering both techniques makes you a more versatile problem solver for JEE Advanced.

Q: Why did we factor out $x^{1/3}$ specifically in the right-hand limit?

We looked at the denominator, which is $x^{4/3}$. By factoring out $x^{1/3}$ from the numerator, we created a term that perfectly cancels out the denominator, leaving us with a much simpler expression. It is all about strategic simplification to make the limit evaluation trivial.

Question

If $f(x) = \begin{cases} \frac{\sin(a+2)x + \sin x}{x}, & x < 0 \ b, & x = 0 \ \frac{(x+3x^2)^{1/3} - x^{1/3}}{x^{4/3}}, & x > 0 \end{cases}$ is continuous at $x = 0$ then $a + 2b$ is equal to:

Options:

(A) $-2$

(B) $1$

(C) $0$

(D) $-1$

Sigma Percentile · Accepted Answer

### The Architecture of Continuity: Building the Bridge

My dear student, welcome to the world of limits. Today, we are not just solving a math problem; we are engineers of a mathematical bridge.

Imagine you are standing at the origin, $x=0$. To your left, there is a path defined by a trigonometric function. To your right, there is a path defined by a complex radical expression. And right beneath your feet, at $x=0$, there is a single point, $b$.

For this bridge to be safe—for the function to be continuous—these three paths must meet at the exact same elevation. Mathematically, this is the sacred condition:

$$\lim_{x 	o 0^-} f(x) = f(0) = \lim_{x 	o 0^+} f(x)$$

Let us build this bridge, step by step.

### Phase 1: The Left-Hand Approach

Let us look to the left. As $x$ approaches $0$ from the negative side, our function is defined as $f(x) = \frac{\sin(a+2)x + \sin x}{x}$. We can split this fraction into two distinct parts:

$$\lim_{x 	o 0^-} \left[ \frac{\sin(a+2)x}{x} + \frac{\sin x}{x} ight]$$

Do you recall the fundamental limit $\lim_{	heta 	o 0} \frac{\sin 	heta}{	heta} = 1$? This is our most powerful tool.

For the first term, we need the denominator to match the argument of the sine function. We multiply and divide by $(a+2)$ to get $\frac{\sin(a+2)x}{(a+2)x} \cdot (a+2)$.

As $x 	o 0$, this term becomes $1 \cdot (a+2) = a+2$. The second term, $\frac{\sin x}{x}$, is the classic limit, which is simply $1$. Adding these together, our Left-Hand Limit (LHL) is $a+3$.

### Phase 2: The Right-Hand Challenge

Now, let us turn to the right. As $x$ approaches $0$ from the positive side, the function is $f(x) = \frac{(x+3x^2)^{1/3} - x^{1/3}}{x^{4/3}}$.

Notice that both terms in the numerator contain $x^{1/3}$. Let us factor it out:

$$\lim_{x 	o 0^+} \frac{x^{1/3}[(1+3x)^{1/3} - 1]}{x^{4/3}}$$

When we divide $x^{1/3}$ by $x^{4/3}$, we subtract the exponents: $\frac{1}{3} - \frac{4}{3} = -1$. This means the $x$ moves to the denominator. Our expression simplifies beautifully to:

$$\lim_{x 	o 0^+} \frac{(1+3x)^{1/3} - 1}{x}$$

Now, we use the binomial approximation. For small $u$, we know that $(1+u)^n \approx 1+nu$. Here, $u = 3x$ and $n = 1/3$.

So, $(1+3x)^{1/3} \approx 1 + \frac{1}{3}(3x) = 1+x$. Substituting this back into our limit, we get:

$$\lim_{x 	o 0^+} \frac{(1+x) - 1}{x} = \lim_{x 	o 0^+} \frac{x}{x} = 1$$

The complex radical expression has melted away, leaving us with a clean, simple $1$. The right side of our bridge is complete.

### Phase 3: The Synthesis

We have our LHL, which is $a+3$. We have our RHL, which is $1$. And we know the value of the function at $x=0$ is $b$. For continuity, all three must be equal:

$$a+3 = b = 1$$

From $b = 1$, we immediately find our first variable. From $a+3 = 1$, we solve for $a$: $a = 1 - 3 = -2$.

The question asks for $a + 2b$. Substituting our values, we get:

$$-2 + 2(1) = -2 + 2 = \mathbf{0}$$

There you have it. Through careful analysis and the power of approximations, we have turned a daunting piecewise function into a simple arithmetic result. Never fear the complexity of a problem; look for the patterns, apply your tools, and build your bridge with confidence.

Animated Solution for Mathematics - Limits, Continuity and Differentiability: If $f (x) = ⎩ ⎨ ⎧ \frac{s i n ( a + 2 ) x + s i n x}{x}, b, \frac{( x + 3 x ^{2} ) ^{1/3} - x ^{1/3}}{x ^{4/3}}, x < 0 x = 0 x > 0$ is continuous at $x = 0$ then $a + 2 b$ is equal to:

Select Answer:

Visualized Solution

Condition for Continuity at $x = 0$

Evaluating Left Hand Limit (LHL)

Splitting the LHL Expression

Applying Standard Limits

Evaluating Right Hand Limit (RHL)

Factoring out $x^{\frac{1}{3}}$

Simplifying the RHL Expression

Applying Binomial Approximation

Finalizing the RHL

Equating Limits for Continuity

Finding the value of $b$

Finding the value of $a$

Calculating $a + 2 b$

The Sigma Insight: Continuity at a Point and in an Interval

The Architecture of Continuity

Phase 1

Phase 2

Phase 3

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