Let e denote the base of the natural logarithm. The value of the real number a for which the right h

Q: Why can't we just use L'Hopital's Rule directly on the original expression?

You could, but it would be a nightmare of differentiation! The chain rule on $(1-x)^{1/x}$ creates a recursive mess. Using the Maclaurin series is the 'sniper' approach—clean, precise, and elegant.

Q: Why do we need the limit to be a non-zero real number?

That condition is the 'key' to the lock. If the limit were zero, $a$ could be anything smaller than the critical value. If it were infinity, $a$ could be anything larger. By demanding a non-zero real value, we force the powers of $x$ to match perfectly, locking $a$ into a single, unique value.

Q: I got confused with the $e^{-1}$ term. Why did we factor it out?

Great observation! We factor out $e^{-1}$ to force the numerator into the form $e^u - 1$. This is the standard gateway to using the limit $\lim_{u \to 0} \frac{e^u - 1}{u} = 1$. Without that $-1$, you cannot apply the standard limit formula.

Question

Let $e$ denote the base of the natural logarithm. The value of the real number $a$ for which the right hand limit $\lim_{x 	o 0^+} \frac{(1 - x)^{1/x} - e^{-1}}{x^a}$ is equal to a nonzero real number, is ______.

Sigma Percentile · Accepted Answer

### The Mystery of the Indeterminate Form

Imagine you are standing before a mathematical gatekeeper. The problem is simple, yet it holds a secret:

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

We are told $L$ is a non-zero real number. Our mission is to find $a$.

At first glance, the term $(1 - x)^{1/x}$ looks innocent, but as $x$ approaches zero, the base approaches $1$ and the exponent approaches $\infty$. This is the classic $1^\infty$ indeterminate form, a siren song that has lured many students into algebraic traps.

### Phase 1: The Base $e$ Transformation

To conquer this, we must bring the variable exponent down to earth. We use the identity $f(x)^{g(x)} = e^{g(x) \ln(f(x))}$.

By applying this to our term, we transform $(1 - x)^{1/x}$ into $e^{\frac{\ln(1 - x)}{x}}$. Now, the expression looks like this:

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

We are getting closer, but that $e^{-1}$ is still blocking our path to the standard limit formula. Let's factor it out.

By pulling $e^{-1}$ out of the numerator, we get $e^{-1} \left( e^{1 + \frac{\ln(1-x)}{x}} - 1 ight)$. Now, look at that exponent: $1 + \frac{\ln(1-x)}{x}$.

If we combine these terms, we get $\frac{x + \ln(1-x)}{x}$. This is the heart of the problem.

### Phase 2: The Maclaurin Series Weapon

We know that for small $u$, $e^u - 1 \approx u$. Let $u = \frac{x + \ln(1-x)}{x}$.

As $x 	o 0$, does $u$ approach zero? Let's check. The Maclaurin series for $\ln(1-x)$ is:

$$\ln(1-x) = -x - \frac{x^2}{2} - \frac{x^3}{3} - \dots$$

Substituting this into our numerator, we get $x + (-x - \frac{x^2}{2} - \frac{x^3}{3} - \dots) = -\frac{x^2}{2} - \frac{x^3}{3} - \dots$.

Since the numerator behaves like $-\frac{x^2}{2}$, the entire exponent $u$ approaches zero as $x 	o 0$. We are cleared for takeoff!

### Phase 3: The Balancing Act

Now we substitute our approximation back into the limit:

$$L = \frac{1}{e} \lim_{x 	o 0^+} \frac{-\frac{x^2}{2} - \frac{x^3}{3} - \dots}{x \cdot x^a}$$

The $x$ from the denominator of the exponent multiplies with $x^a$ to become $x^{a+1}$. So, we have:

$$L = \frac{1}{e} \lim_{x 	o 0^+} \frac{-\frac{x^2}{2} - \frac{x^3}{3} - \dots}{x^{a+1}}$$

For this limit to be a non-zero real number, the power of $x$ in the numerator must perfectly match the power of $x$ in the denominator. The lowest power in our numerator is $2$.

Therefore, we must have $a + 1 = 2$. Solving this gives us **$a = 1$**.

### Conclusion

And there it is! By carefully peeling back the layers of the exponential function and using the Maclaurin series to expose the true behavior of the numerator, we found that **$a = 1$**.

It is a beautiful result—simple, elegant, and deeply satisfying. Remember, in limits, it is never about brute force; it is about finding the dominant term that dictates the behavior of the function.

Enter Numerical Value:

Visualized Solution

The Limit Problem

Converting to Base $e$

Factoring out $e^{- 1}$

Simplifying the Exponent

Standard Exponential Limit

Applying the Approximation

Maclaurin Series Expansion

Simplifying the Numerator

Condition for Non-Zero Finite Limit

Solving for $a$

The Sigma Insight: Evaluation of Limits & L'Hopital's Rule

The Mystery of the Indeterminate Form

Phase 1

Phase 2

Phase 3

Conclusion

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Enter Numerical Value:

Visualized Solution

The Limit Problem

Converting to Base .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } e

Factoring out .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } e−1