We are tasked with evaluating the limit:

$\lim_{{x \to 1}} \ln\left( \frac{x}{e^{8x}} \right)$

Step 1: Simplify the argument of the logarithm

First, rewrite the expression inside the logarithm:

$\ln\left( \frac{x}{e^{8x}} \right) = \ln(x) - \ln(e^{8x})$

Since $\ln(e^{8x}) = 8x$ (by the logarithmic identity $\ln(e^a) = a$):

$\ln(x) - 8x$

Thus, the limit becomes:

$\lim_{{x \to 1}} \left( \ln(x) - 8x \right)$

Step 2: Evaluate the limit

Now, evaluate each term at $x = 1$:

• $\lim_{{x \to 1}} \ln(x) = \ln(1) = 0$
• $\lim_{{x \to 1}} 8x = 8(1) = 8$

Therefore:

$\lim_{{x \to 1}} \left( \ln(x) - 8x \right) = 0 - 8 = -8$

Conclusion:

The limit is:

$\boxed{-8}$

Would you like further details on this solution or have any questions?

Follow-up questions:

1. How does the natural logarithmic function behave as $x \to 0$ or $x \to \infty$?
2. Can you explain the properties of logarithms that were used in the simplification process?
3. What is the significance of exponential growth in the expression $e^{8x}$?
4. How would this problem change if the exponent of $e$ was different?
5. How do logarithms and exponentials interact in calculus and limits?

Tip:

When simplifying logarithmic expressions, always remember to apply logarithmic identities like $\ln(a^b) = b \ln(a)$ and $\ln(ab) = \ln(a) + \ln(b)$.