If f(x) = { (sin(p+1)x + sin x)/x, x 0 } is continuous at

Q: Wait, why can't I just plug in $x=0$ directly into the RHL expression?

If you plug in $x=0$ directly, you get $\frac{\sqrt{0+0} - \sqrt{0}}{0^{3/2}}$, which is $\frac{0}{0}$. This is an indeterminate form. In calculus, this tells us that the function is hiding its true value behind a 'hole' or a point of discontinuity. We must simplify the expression algebraically to reveal the limit's true value.

Q: How do I know when to use the binomial expansion vs. rationalization?

Both are valid! Rationalization is great when you have square roots like $\sqrt{A} - \sqrt{B}$. Binomial expansion, specifically $(1+x)^n \approx 1 + nx$, is a powerful shortcut for limits as $x \to 0$. In this problem, either method will lead you to the same result of $1/2$. Practice both to see which feels more intuitive for you.

Q: Why is it necessary to check the LHL and RHL separately?

Because the function is defined piecewise! The rule for the function changes depending on whether $x$ is negative or positive. The only way to ensure the 'bridge' is connected is to verify that the limit from the left (using the $x 0$ rule) both meet at the same value $q$.

Question

If $f(x) = \begin{cases} \frac{\sin(p+1)x + \sin x}{x} & x < 0 \ q & x = 0 \ \frac{\sqrt{x+x^2} - \sqrt{x}}{x^{3/2}} & x > 0 \end{cases}$ is continuous at $x = 0$, then the ordered pair $(p,q)$ is equal to :

Options:

(A) $(5/2, 1/2)$

(B) $(-3/2, 1/2)$

(C) $(1/2, 3/2)$

(D) $(-3/2, 1/2)$

Sigma Percentile · Accepted Answer

### The Architecture of Continuity: Building the Perfect Bridge

Imagine you are an engineer tasked with connecting two roads. One road comes from the left, another from the right, and they must meet at a specific point, $x = 0$.

If they don't meet at the exact same height, your bridge is broken, and the function is discontinuous. In the world of JEE Advanced, continuity is simply the mathematical guarantee that this bridge is perfectly seamless.

Our function $f(x)$ is defined in three pieces. To ensure continuity at $x = 0$, we must satisfy the condition: 
$$\lim_{x 	o 0^-} f(x) = f(0) = \lim_{x 	o 0^+} f(x)$$

We are given $f(0) = q$. This $q$ is the height of our central pillar. Let's find the heights of the roads approaching it.

### Phase 1: The Right Hand Limit (The Algebraic Challenge)

Let's look at the right side, where $x > 0$. The function is defined as $f(x) = \frac{\sqrt{x+x^2} - \sqrt{x}}{x^{3/2}}$. As $x$ approaches $0$, this expression results in the indeterminate form $0/0$.

Look at the numerator: $\sqrt{x+x^2}$. We can factor out an $x$ to get $\sqrt{x(1+x)}$, which is $\sqrt{x}\sqrt{1+x}$. Now, the numerator becomes $\sqrt{x}\sqrt{1+x} - \sqrt{x}$.

If we factor out $\sqrt{x}$, we get $\sqrt{x}(\sqrt{1+x} - 1)$. Now, look at the denominator: $x^{3/2}$. We can write this as $x \cdot \sqrt{x}$.

Putting it all together, our limit becomes:

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

The $\sqrt{x}$ terms cancel out beautifully. We are left with $\lim_{x 	o 0^+} \frac{\sqrt{1+x} - 1}{x}$.

Using the standard limit $\lim_{x 	o 0} \frac{(1+x)^n - 1}{x} = n$, where $n = 1/2$, we find that the Right Hand Limit is exactly $1/2$. Our right-side road is at height $1/2$. Thus, $q = 1/2$.

### Phase 2: The Left Hand Limit (The Trigonometric Challenge)

Now, let's turn to the left side, where $x < 0$. The function is $f(x) = \frac{\sin(p+1)x + \sin x}{x}$. We need to find the limit as $x 	o 0^-$.

We can split this into two separate limits:

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

Recall the standard trigonometric limit: $\lim_{	heta 	o 0} \frac{\sin(a	heta)}{	heta} = a$. Applying this to our terms, the first limit becomes $(p+1)$ and the second limit becomes $1$.

Adding them together, the Left Hand Limit is $p + 2$.

### Phase 3: The Synthesis

For the bridge to be continuous, the Left Hand Limit must equal the function's value at zero, which we already determined is $q = 1/2$. So, we set up the final equation:

$$p + 2 = q$$

Substituting $q = 1/2$ into the equation, we get $p + 2 = 1/2$. Solving for $p$, we find $p = 1/2 - 2 = -3/2$.

We have arrived at our destination! The ordered pair $(p, q)$ is **$(-3/2, 1/2)$**.

You have successfully navigated the piecewise definition, conquered the indeterminate algebraic form, and mastered the trigonometric limits. This is the essence of JEE Advanced—not just solving, but understanding the structural integrity of the math itself.

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

Select Answer:

Visualized Solution

The Concept of Continuity at $x = 0$

The Continuity Condition

Setting up the Right Hand Limit

Factoring the Numerator

Simplifying the Expression

Evaluating the Standard Limit

Finding the Value of $q$

Setting up the Left Hand Limit

Splitting the Limit

Applying Standard Trig Limits

Simplifying the LHL

Equating LHL and $f (0)$

Solving for $p$

The Final Ordered Pair

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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