Solution:

i. $\lim_{x \to 2} \frac{\sqrt{x^2 + 4x - 3}}{x - 3}$

Step 1: Direct substitution check
Substituting $x = 2$: $\sqrt{2^2 + 4(2) - 3} = \sqrt{4 + 8 - 3} = \sqrt{9} = 3$ And the denominator: $x - 3 = 2 - 3 = -1$ So, direct substitution gives: $\frac{3}{-1} = -3$

Final Answer for i: $-3$

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ii. $\lim_{x \to 1} \frac{|x - 2|}{x - 2}$

Step 1: Analyze the absolute value
The expression $|x - 2|$ depends on the value of $x$ relative to 2:

• If $x > 2$, $|x - 2| = x - 2$
• If $x < 2$, $|x - 2| = -(x - 2)$

Here, $x$ approaches $1$, so $x < 2$, and: $|x - 2| = -(x - 2)$

Thus, the fraction becomes: $\frac{|x - 2|}{x - 2} = \frac{-(x - 2)}{x - 2} = -1$

Step 2: Verify behavior as $x \to 1$
Since $x$ approaches $1$, the sign of $x - 2$ remains negative, confirming the above result.

Final Answer for ii: $-1$

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Summary of Results:

1. $\lim_{x \to 2} \frac{\sqrt{x^2 + 4x - 3}}{x - 3} = -3$
2. $\lim_{x \to 1} \frac{|x - 2|}{x - 2} = -1$

Would you like detailed explanations or derivations for either part?
Here are 5 related questions to deepen understanding:

1. How do you evaluate limits involving absolute values approaching critical points?
2. What are common techniques to resolve indeterminate forms in limits?
3. How does the behavior of a function differ when approaching a limit from the left vs. right?
4. Can you explain the geometric interpretation of limits for these problems?
5. What are practical applications of limits in real-world scenarios?

Tip: When working with limits involving absolute values, always break them into piecewise functions to analyze behavior in different intervals.