The given limit is:

$\lim_{x \to 0} (1 - 4x)^{1/x}.$

To solve this, we can rewrite the expression in a more convenient logarithmic form:

Step 1: Rewrite Using Logarithms

Let $y = (1 - 4x)^{1/x}$. Taking the natural logarithm of both sides: $\ln y = \frac{1}{x} \ln(1 - 4x).$

So, the limit becomes: $\lim_{x \to 0} \ln y = \lim_{x \to 0} \frac{\ln(1 - 4x)}{x}.$

Step 2: Apply l'Hospital's Rule

This limit is an indeterminate form ($\frac{0}{0}$), so we apply l'Hospital's Rule by differentiating the numerator and denominator:

• Derivative of the numerator $\ln(1 - 4x)$: $\frac{-4}{1 - 4x}$,
• Derivative of the denominator $x$: $1$.

The limit now becomes: $\lim_{x \to 0} \frac{\ln(1 - 4x)}{x} = \lim_{x \to 0} \frac{-4}{1 - 4x}.$

Step 3: Evaluate the Limit

As $x \to 0$, $1 - 4x \to 1$. Thus: $\lim_{x \to 0} \frac{-4}{1 - 4x} = -4.$

So, $\lim_{x \to 0} \ln y = -4$. Exponentiating both sides to get $y$, we find: $y = e^{-4}.$

Final Answer:

$\lim_{x \to 0} (1 - 4x)^{1/x} = e^{-4}.$

Would you like a step-by-step explanation of any part? Let me know!

Related Questions:

1. How do you determine when to use logarithmic transformations in limits?
2. What are the conditions for applying l'Hospital's Rule?
3. How do exponential functions behave under small changes in their base?
4. Why is $e^x$ significant in limits involving indeterminate forms?
5. Can this limit be solved without logarithms? Why or why not?

Tip: Always check for indeterminate forms ($0/0$ or $\infty/\infty$) before applying l'Hospital's Rule.