Let f : R → R be a function defined as f(x) = { (sin(a+1)x + sin 2x)/2x, if x

Q: Wait, why can't I just plug in $x=0$ immediately for the right-hand limit?

That is a classic trap! If you plug in $x=0$ directly, you get $\frac{\sqrt{0} - \sqrt{0}}{b(0)^{5/2}}$, which is $\frac{0}{0}$. This is an indeterminate form, meaning the limit exists, but it is hidden. You must simplify the expression algebraically first to reveal the true value.

Q: Why did the $bx^3$ term in the RHL simplify so cleanly?

It is not magic, it is algebra! When you rationalize the numerator, you get $(x+bx^3) - x$, which is $bx^3$. In the denominator, you have $bx^{5/2}$ multiplied by the conjugate. When you factor $\sqrt{x}$ out of the conjugate, you get $bx^{5/2} \cdot \sqrt{x} = bx^3$. The $bx^3$ terms in the numerator and denominator then cancel out perfectly, leaving you with a simple limit.

Q: Is $b$ really just a constant?

Yes! In the definition of the function, $f(0) = b$. Since the function is continuous at $x=0$, the value of the function at that point must be equal to the limits from both sides. So, $b$ is the anchor that ties the left-hand limit and the right-hand limit together.

Question

Let $f : \mathbb{R} 	o \mathbb{R}$ be a function defined as $f(x) = \begin{cases} \frac{\sin(a+1)x + \sin 2x}{2x}, & 	ext{if } x < 0 \ b, & 	ext{if } x = 0 \ \frac{\sqrt{x+bx^3} - \sqrt{x}}{bx^{5/2}}, & 	ext{if } x > 0 \end{cases}$ If $f$ is continuous at $x = 0$, then the value of $a + b$ is equal to:

Options:

(A) $-\frac{5}{2}$

(B) -2

(C) -3

(D) $-\frac{3}{2}$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

In the world of calculus, continuity represents the structural integrity of a function. For a function $f(x)$ to be continuous at $x = 0$, the bridge must be unbroken, satisfying the condition:

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

This equality ensures that the left-hand limit (LHL), the right-hand limit (RHL), and the function value at the point are perfectly aligned.

### Phase 1: The Left-Hand Journey

We approach $x = 0$ from the left side ($x < 0$), where the function is defined as:

$$f(x) = \frac{\sin(a+1)x + \sin 2x}{2x}$$

To evaluate this limit, we split the expression into two distinct parts:

$$\lim_{x 	o 0^-} \left( \frac{\sin(a+1)x}{2x} + \frac{\sin 2x}{2x} ight)$$

Recalling the standard limit $\lim_{	heta 	o 0} \frac{\sin k	heta}{	heta} = k$, we apply this to our terms. The first part evaluates to $\frac{a+1}{2}$ and the second part evaluates to $\frac{2}{2} = 1$.

Summing these results, the height of our bridge from the left is:

$$\frac{a+1}{2} + 1 = \frac{a+3}{2}$$

### Phase 2: The Right-Hand Challenge

Now, we approach from the right ($x > 0$), where the function is:

$$f(x) = \frac{\sqrt{x+bx^3} - \sqrt{x}}{bx^{5/2}}$$

Direct substitution yields an indeterminate $\frac{0}{0}$ form, necessitating rationalization. Multiplying the numerator and denominator by the conjugate $\sqrt{x+bx^3} + \sqrt{x}$, the numerator simplifies to $(x+bx^3) - x = bx^3$.

In the denominator, we have $bx^{5/2}(\sqrt{x+bx^3} + \sqrt{x})$. Factoring $\sqrt{x}$ out of the bracket, we obtain:

$$bx^{5/2} \cdot \sqrt{x}(\sqrt{1+bx^2} + 1) = bx^3(\sqrt{1+bx^2} + 1)$$

The $bx^3$ terms cancel, leaving:

$$\lim_{x 	o 0^+} \frac{1}{\sqrt{1+bx^2} + 1} = \frac{1}{1+1} = \frac{1}{2}$$

### Phase 3: The Synthesis

For the function to be continuous, the LHL, RHL, and $f(0)$ must be equal. Given $f(0) = b$, we set up the following equality:

$$\frac{a+3}{2} = b = \frac{1}{2}$$

From this, we immediately identify that **$b = \frac{1}{2}$**. Setting the LHL equal to the RHL:

$$\frac{a+3}{2} = \frac{1}{2} \Rightarrow a+3 = 1 \Rightarrow a = -2$$

Finally, we calculate the requested sum $a+b$:

$$-2 + \frac{1}{2} = -\frac{3}{2}$$

The final result is **$-\frac{3}{2}$**.

Select Answer:

Visualized Solution

Defining Continuity at $x = 0$

The Continuity Equation

Evaluating the Left Hand Limit

Applying the Standard Sine Limit

Simplifying the $L H L$ Expression

Evaluating the Right Hand Limit

Rationalizing the Expression

Algebraic Simplification of $R H L$

Final $R H L$ Value

Equating the Limits and $f (0)$

Solving for Parameters $a$ and $b$

Calculating the Final Sum $a + b$

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

Analyzing the Setup

Phase 1

Phase 2

Phase 3

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Select Answer:

Visualized Solution

Defining Continuity at .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } x=0

The Continuity Equation

Evaluating the Left Hand Limit

Applying the Standard Sine Limit

Simplifying the .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } LHL Expression

Evaluating the Right Hand Limit

Rationalizing the Expression

Algebraic Simplification of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } RHL

Final .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } RHL Value

Equating the Limits and .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } f(0)

Solving for Parameters .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } a and .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } b

Calculating the Final Sum .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } a+b