Let f(x) = {(1 + ax)^(1/x), x 0} be continuo

Q: Wait, why do we reject the negative root when solving for $c$?

In this problem, we are dealing with a cube root function $\sqrt[3]{x+c}$. When we set the denominator to zero at $x=0$, we get $c^{1/3} - 2 = 0$. Since $2$ is the unique real cube root of $8$, $c$ must be $8$. There is no ambiguity here because the cube root function is one-to-one, unlike square roots which can have positive and negative branches.

Q: Why can't we just use direct substitution for the right-hand limit?

If you try direct substitution, you get $\frac{\sqrt{4}-2}{\sqrt[3]{c}-2} = \frac{0}{\sqrt[3]{c}-2}$. If the denominator were not zero, the limit would be $0$. However, for the function to be continuous, the limit must match the other parts, which we find are non-zero. Thus, we must force the denominator to be zero to create a $\frac{0}{0}$ indeterminate form, which allows the limit to potentially be a non-zero finite value.

Q: Is L'Hôpital's Rule the only way to solve the right-hand limit?

Not at all! You could also use rationalization or binomial expansion. For example, you could multiply the numerator and denominator by the appropriate conjugate or use the expansion $(1+u)^n \approx 1+nu$. L'Hôpital's Rule is just the most direct path here because the derivatives of these roots are straightforward.

Question

Let $f(x) = \begin{cases} (1 + ax)^{1/x} & , x < 0 \ 1 + b & , x = 0 \ \frac{(x+4)^{1/2}-2}{(x+c)^{1/3}-2} & , x > 0 \end{cases}$ be continuous at $x = 0$. Then $e^a bc$ is equal to

Options:

(A) 64

(B) 72

(C) 48

(D) 36

Sigma Percentile · Accepted Answer

### The Art of Unbroken Paths: Mastering Continuity

Imagine you are an artist sketching a continuous curve on a canvas. The golden rule of continuity is simple: you must never lift your pen.

In the world of calculus, this means that as you approach a point from the left, arrive at the point itself, and then depart toward the right, the path must be seamless. This is exactly what we are testing with our function $f(x)$.

We are given a piecewise function and told it is continuous at $x = 0$. This is our anchor. It tells us that the Left-Hand Limit (LHL), the function value at $x = 0$, and the Right-Hand Limit (RHL) must all be equal.

Let us embark on this journey to find the hidden values of $a$, $b$, and $c$.

### Phase 1: The Left-Hand Limit and the $1^\infty$ Mystery

We begin by looking at the region where $x < 0$. Here, $f(x) = (1 + ax)^{1/x}$.

As $x$ creeps closer to $0$ from the negative side, the base $(1 + ax)$ approaches $1$, and the exponent $1/x$ shoots off toward $-\infty$. This is the classic $1^\infty$ indeterminate form.

We have a powerful tool for this: the exponential limit formula. We know that:

$$\lim_{x 	o 0} (1 + f(x))^{g(x)} = e^{\lim_{x 	o 0} f(x)g(x)}$$

Applying this to our function, we get $e^{\lim_{x 	o 0} (ax \cdot \frac{1}{x})}$. The $x$ terms cancel out with elegant precision, leaving us with $e^a$. This is our LHL.

### Phase 2: The Right-Hand Limit and the Zero-Denominator Trap

Now, we shift our focus to the right side, where $x > 0$. Our function is:

$$f(x) = \frac{(x+4)^{1/2}-2}{(x+c)^{1/3}-2}$$

As $x 	o 0^+$, the numerator becomes $\sqrt{4} - 2 = 0$. If the denominator were any non-zero number, the limit would be $0$.

But we know the limit must be equal to our LHL, which is $e^a$ (and $e^a$ is definitely not $0$). Therefore, the denominator MUST also approach $0$ to create a $\frac{0}{0}$ indeterminate form.

This forces the condition $\sqrt[3]{0+c} - 2 = 0$, which simplifies to $c^{1/3} = 2$. Cubing both sides, we find **$c = 8$**.

### Phase 3: The Final Synthesis

With $c = 8$, our RHL becomes:

$$\lim_{x 	o 0^+} \frac{\sqrt{x+4}-2}{\sqrt[3]{x+8}-2}$$

This is now a perfect $\frac{0}{0}$ form. We apply L'Hôpital's Rule, differentiating the numerator and denominator with respect to $x$.

The derivative of the numerator is $\frac{1}{2}(x+4)^{-1/2}$, and the derivative of the denominator is $\frac{1}{3}(x+8)^{-2/3}$. Evaluating these at $x = 0$, we get:

$$\frac{1/4}{1/12} = 3$$

Now, we bring it all together. We have:

LHL $= e^a$

Value $= 1 + b$

RHL $= 3$

Since the function is continuous, $e^a = 1 + b = 3$. This gives us $e^a = 3$ and $b = 3 - 1 = 2$.

We have all our variables: $e^a = 3$, $b = 2$, and $c = 8$. The final step is to calculate the product:

$$e^a bc = 3 \cdot 2 \cdot 8 = 48$$

We have successfully navigated the constraints of continuity to reach our destination. The final answer is **$48$**.

Animated Solution for Mathematics - Limits, Continuity and Differentiability: Let $f (x) = ⎩ ⎨ ⎧ (1 + a x)^{1/ x} 1 + b \frac{( x + 4 ) ^{1/2} - 2}{( x + c ) ^{1/3} - 2}, x < 0, x = 0, x > 0$ be continuous at $x = 0$ . Then $e^{a} b c$ is equal to

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Visualized Solution

Understanding Continuity

The Continuity Equation

Setting up the Left-Hand Limit (LHL)

Evaluating the LHL

The Function Value at $x = 0$

Setting up the Right-Hand Limit (RHL)

The Zero Denominator Condition

Finding the Value of $c$

Applying L'Hôpital's Rule

Calculating the RHL Value

Equating the Limits

Solving for $e^{a}$ and $b$

Final Calculation: $e^{a} b c$

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

The Art of Unbroken Paths

Phase 1

Phase 2

Phase 3

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