Let's solve each of these limits step-by-step.

Problem (i)

$\lim_{{x \to 0}} \frac{|x|}{x}$

This limit depends on the direction from which $x$ approaches 0, due to the absolute value in the numerator.

1. As $x \to 0^+$ (from the right): $|x| = x$, so: $\frac{|x|}{x} = \frac{x}{x} = 1$ Therefore, $\lim_{{x \to 0^+}} \frac{|x|}{x} = 1$

2. As $x \to 0^-$ (from the left): $|x| = -x$, so: $\frac{|x|}{x} = \frac{-x}{x} = -1$ Therefore, $\lim_{{x \to 0^-}} \frac{|x|}{x} = -1$

Since the left-hand limit and right-hand limit are not equal, the two-sided limit does not exist: $\lim_{{x \to 0}} \frac{|x|}{x} \text{ does not exist}$

Problem (ii)

$\lim_{{x \to 0}} \frac{5^x - 1}{3^x - 1}$

To evaluate this, we can use the fact that for any constant $a$, $\lim_{{x \to 0}} \frac{a^x - 1}{x} = \ln(a)$. Therefore, we can rewrite this limit as:

$\lim_{{x \to 0}} \frac{5^x - 1}{3^x - 1} = \frac{\lim_{{x \to 0}} \frac{5^x - 1}{x}}{\lim_{{x \to 0}} \frac{3^x - 1}{x}}$

Using the property mentioned: $\lim_{{x \to 0}} \frac{5^x - 1}{x} = \ln(5) \quad \text{and} \quad \lim_{{x \to 0}} \frac{3^x - 1}{x} = \ln(3)$

Thus, $\lim_{{x \to 0}} \frac{5^x - 1}{3^x - 1} = \frac{\ln(5)}{\ln(3)}$

Final Answers

1. (i) $\lim_{{x \to 0}} \frac{|x|}{x}$: The limit does not exist.
2. (ii) $\lim_{{x \to 0}} \frac{5^x - 1}{3^x - 1} = \frac{\ln(5)}{\ln(3)}$

Would you like more details on any part of the solution?

Related Questions

1. What is the interpretation of left-hand and right-hand limits?
2. How do we evaluate limits involving exponential functions without L'Hôpital's rule?
3. Can we use L'Hôpital's rule directly on $\frac{5^x - 1}{3^x - 1}$?
4. What are some properties of logarithmic functions used in limits?
5. How do limits of absolute values affect the existence of two-sided limits?

Tip

When dealing with limits that involve different behaviors from the left and right, always check both one-sided limits separately.