For the function f(x) = x / (1 + e^(1/x)), x ≠ 0; = 0, x = 0, the derivative from the right, f'(0+)

Q: Wait, why do we use $f(0)=0$ in the limit definition? Is that just given?

Yes, exactly! In a piecewise function, the value at the point of interest is explicitly defined. Here, the problem states $f(0) = 0$. When we calculate the derivative at $x=0$, we must use this specific value in the definition $\lim_{h \to 0} \frac{f(0+h) - f(0)}{h}$.

Q: Why does $e^{1/h}$ go to infinity when $h \to 0^+$?

Think about the exponent. As $h$ approaches $0$ from the positive side, $h$ is a very small positive number like $0.0001$. Therefore, $1/h$ becomes a very large positive number like $10,000$. As the exponent $1/h$ grows towards infinity, $e^{1/h}$ also grows towards infinity. It is the fundamental behavior of the exponential function!

Q: Can I just use the quotient rule to differentiate $f(x)$?

That is a common trap! The quotient rule is valid only where the function is already differentiable. Since we are testing for differentiability at $x=0$, we cannot assume the derivative exists. We must use the first-principles definition of the derivative to check if the left-hand and right-hand limits of the slope are equal.

Question

For the function $f(x) = \begin{cases} \frac{x}{1 + e^{1/x}}, & x 
eq 0 \ 0, & x = 0 \end{cases}$, the derivative from the right, $f'(0^+) = .........$, and the derivative from the left, $f'(0^-) = .........$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Welcome, my dear student, to a fascinating exploration of calculus. Today, we are not just solving a problem; we are dissecting the very nature of smoothness in functions.

We are looking at the function defined as:

$$f(x) = \begin{cases} \frac{x}{1 + e^{1/x}}, & x 
eq 0 \ 0, & x = 0 \end{cases}$$

At first glance, it looks like a standard rational function, but the presence of the exponential term $e^{1/x}$ introduces a fascinating behavior near the origin. Our mission is to determine if this function is "smooth" at $x=0$ by calculating its right-hand and left-hand derivatives.

### The Right-Hand Approach (The Vanishing Slope)

Imagine you are walking along the curve from the positive side, moving towards the origin. We want to find the slope of the tangent line as we approach $x=0$ from the right.

We use the definition of the derivative:

$$f'(0^+) = \lim_{h 	o 0^+} \frac{f(0+h) - f(0)}{h}$$

Substituting our function, we get:

$$f'(0^+) = \lim_{h 	o 0^+} \frac{\frac{h}{1 + e^{1/h}} - 0}{h}$$

Notice how the $h$ in the numerator and the $h$ in the denominator cancel out perfectly? This leaves us with:

$$f'(0^+) = \lim_{h 	o 0^+} \frac{1}{1 + e^{1/h}}$$

Now, let us apply the limit. As $h 	o 0^+$, the exponent $1/h$ shoots off to positive infinity. Consequently, $e^{1/h}$ also explodes towards infinity.

Our expression becomes $\frac{1}{1 + \infty}$, which is $0$. The slope from the right is perfectly horizontal!

### The Left-Hand Approach (The Unit Slope)

Now, let us switch our perspective. We approach the origin from the negative side. The formula changes slightly:

$$f'(0^-) = \lim_{h 	o 0^+} \frac{f(0-h) - f(0)}{-h}$$

Substituting $x = -h$ into our function, we get $f(-h) = \frac{-h}{1 + e^{-1/h}}$. Plugging this into our limit, we have:

$$f'(0^-) = \lim_{h 	o 0^+} \frac{\frac{-h}{1 + e^{-1/h}} - 0}{-h}$$

Again, the $-h$ terms cancel out, leaving us with:

$$f'(0^-) = \lim_{h 	o 0^+} \frac{1}{1 + e^{-1/h}}$$

This is where the magic happens. As $h 	o 0^+$, the exponent $-1/h$ approaches negative infinity. What happens to $e$ raised to a massive negative power? It vanishes to $0$!

Thus, our expression simplifies to $\frac{1}{1 + 0}$, which is $1$.

### The Conclusion: A Sharp Encounter

We have found that the slope from the right is $0$, while the slope from the left is $1$. Because these two slopes are not equal, the function does not have a single, defined tangent at the origin.

Instead, it possesses a **"sharp corner."** In the language of calculus, this means the function is **not differentiable at $x=0$**.

I hope this journey through the limits has illuminated the beauty of piecewise analysis. Keep questioning, keep visualizing, and most importantly, keep falling in love with the logic of mathematics!

Animated Solution for Mathematics - Limits, Continuity and Differentiability: For the function $f (x) = {\frac{x}{1 + e ^{1/ x}}, 0, x \neq = 0 x = 0$ , the derivative from the right, $f^{'} (0^{+}) = .........$ , and the derivative from the left, $f^{'} (0^{-}) = .........$

Visualized Solution

Visualizing the Function $f (x)$

Right-Hand Derivative (RHD) Formula

Substitution for $f^{'} (0^{+})$

Simplifying the RHD Expression

Evaluating the RHD Limit

Left-Hand Derivative (LHD) Formula

Substitution for $f^{'} (0^{-})$

Simplifying the LHD Expression

Evaluating the LHD Limit

Final Conclusion

The Sigma Insight: Differentiability of a Function

Analyzing the Setup

The Right-Hand Approach (The Vanishing Slope)

The Left-Hand Approach (The Unit Slope)

The Conclusion

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Visualized Solution

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