For a, b 0, let f(x) = { (tan((a+1)x) + b tan x)/x, x

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

Plugging in $x=0$ directly gives you an indeterminate form like $0/0$. In calculus, this is a signal that the function is 'hiding' its true value. We use limits to see what the function is approaching as it gets infinitely close to zero, bypassing the undefined point.

Q: Why do we rationalize the numerator in the RHL?

When you see expressions like $\sqrt{A} - \sqrt{B}$, the difference of square roots often prevents us from seeing the common factors that cause the $0/0$ form. Rationalization transforms the difference into a product, allowing those problematic terms to cancel out.

Q: Is it possible for the LHL and RHL to be different?

If the LHL and RHL were different, the function would have a 'jump discontinuity' at $x=0$. Since the problem explicitly states the function is continuous, we are guaranteed that the LHL, RHL, and $f(0)$ are all equal. This is the key constraint that allows us to solve for $a$ and $b$.

Question

For $a, b > 0$, let $f(x) = \begin{cases} \frac{	an((a+1)x) + b 	an x}{x}, & x < 0 \ 3, & x = 0 \ \frac{\sqrt{ax + b^2x^2} - \sqrt{ax}}{b\sqrt{ax}x}, & x > 0 \end{cases}$ be a continuous function at $x = 0$. Then $\frac{b}{a}$ is equal to :

Options:

(A) 6

(B) 4

(C) 5

(D) 8

Sigma Percentile · Accepted Answer

### The Beauty of the Bridge: Understanding Continuity

Welcome, fellow traveler on the JEE journey. Today, we are going to dissect a problem that, at first glance, might seem like a chaotic mess of square roots and trigonometric functions. In the world of calculus, continuity is not just a definition; it is a bridge.

When we say a function is continuous at a point, we are saying that the path from the left and the path from the right meet perfectly at a single, solid destination. There is no jump, no gap, no uncertainty.

Our function $f(x)$ is defined in three pieces. At $x=0$, the value is fixed at $3$. For the function to be continuous, the limit as $x$ approaches $0$ from the left (the LHL) must be $3$, and the limit as $x$ approaches $0$ from the right (the RHL) must also be $3$.

### Phase 1: The Left Bank — Taming the Trigonometry

Let's look at the left side, where $x < 0$. Our function is defined as $f(x) = \frac{	an((a+1)x) + b 	an x}{x}$. When we approach zero, we are dealing with the classic indeterminate form $\frac{0}{0}$.

We have a powerful tool in our arsenal: the standard limit $\lim_{x 	o 0} \frac{	an(kx)}{x} = k$. Instead of panicking, let's split the fraction. We can write the limit as:

$$	ext{LHL} = \lim_{x 	o 0^-} \frac{	an((a+1)x)}{x} + \lim_{x 	o 0^-} \frac{b 	an x}{x}$$

Look at how elegant this becomes! For the first term, our constant $k$ is $(a+1)$. By the standard limit property, this term simply becomes $(a+1)$.

For the second term, the constant $b$ sits outside, and the limit of $\frac{	an x}{x}$ is just $1$. Thus, our LHL simplifies beautifully to $a + 1 + b$. Since we know the LHL must equal $f(0) = 3$, we arrive at our first vital equation:

$$a + b + 1 = 3 \Rightarrow a + b = 2$$

Keep this safe; it is the foundation of our solution.

### Phase 2: The Right Bank — The Art of Rationalization

Now, let's cross over to the right side, where $x > 0$. The function here is $f(x) = \frac{\sqrt{ax + b^2x^2} - \sqrt{ax}}{b\sqrt{ax}x}$. If you try to plug in $x=0$, you get $\frac{0}{0}$.

Whenever you see a difference of square roots, your instinct should be to rationalize the numerator. We multiply the numerator and the denominator by the conjugate: $(\sqrt{ax + b^2x^2} + \sqrt{ax})$.

The numerator transforms into a difference of squares:

$$(\sqrt{ax + b^2x^2})^2 - (\sqrt{ax})^2 = (ax + b^2x^2) - ax = b^2x^2$$

See that? The $ax$ terms vanish! The complexity is melting away. Now, let's look at the denominator. We have $b\sqrt{ax}x$ multiplied by our conjugate. By factoring $\sqrt{x}$ out of the square root terms, we can simplify the denominator to $b\sqrt{a}x^2(\sqrt{a + b^2x} + \sqrt{a})$.

### Phase 3: The Synthesis — Bringing it All Together

Now, watch the magic happen. When we put the simplified numerator and denominator back together, the $x^2$ terms cancel out completely. We are left with:

$$	ext{RHL} = \lim_{x 	o 0^+} \frac{b^2x^2}{b\sqrt{a}x^2(\sqrt{a + b^2x} + \sqrt{a})} = \frac{b}{\sqrt{a}(\sqrt{a} + \sqrt{a})} = \frac{b}{2a}$$

We have successfully navigated the storm. The RHL is $\frac{b}{2a}$. Since the function is continuous, this must also equal $3$.

Therefore, $\frac{b}{2a} = 3$, which gives us $b = 6a$. We now have a system of two simple linear equations:

1. $a + b = 2$
2. $b = 6a$

Substituting the second into the first, we get $a + 6a = 2$, which means $7a = 2$, so $a = \frac{2}{7}$. Consequently, $b = 6 	imes \frac{2}{7} = \frac{12}{7}$.

The question asks for the ratio $\frac{b}{a}$. Dividing $\frac{12}{7}$ by $\frac{2}{7}$ gives us exactly **$6$**.

Animated Solution for Mathematics - Limits, Continuity and Differentiability: For $a, b > 0$ , let $f (x) = ⎩ ⎨ ⎧ \frac{t a n (( a + 1 ) x ) + b t a n x}{x}, 3, \frac{a x + b ^{2} x ^{2} - a x}{b a x x}, x < 0 x = 0 x > 0$ be a continuous function at $x = 0$ . Then $\frac{b}{a}$ is equal to :

Select Answer:

Visualized Solution

Condition for Continuity at $x = 0$

Setting up the Left Hand Limit (LHL)

Evaluating the LHL

Equating LHL to $f (0)$

Setting up the Right Hand Limit (RHL)

Rationalizing the RHL Numerator

Simplifying the RHL Denominator

Final Evaluation of RHL

Equating RHL to $f (0)$

Solving for $a$ and $b$

Final Answer: Finding $\frac{b}{a}$

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

The Beauty of the Bridge

Phase 1

Phase 2

Phase 3

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