(a) lim h→0 √4+h−2 h        (b) lim x→4 x2−16 √x−2      solove using algebraic simplification

Let's solve both limits using algebraic simplification.

### (a) \(\lim_{h \to 0} \frac{\sqrt{4+h} - 2}{h}\)

1. **Multiply by the conjugate:** To simplify, we multiply both the numerator and the denominator by the conjugate of the numerator, \(\sqrt{4+h} + 2\).
   \[
   \lim_{h \to 0} \frac{\sqrt{4+h} - 2}{h} \times \frac{\sqrt{4+h} + 2}{\sqrt{4+h} + 2}
   \]
   This results in:
   \[
   \lim_{h \to 0} \frac{(\sqrt{4+h})^2 - 2^2}{h (\sqrt{4+h} + 2)}
   \]
   Simplifying the numerator:
   \[
   \lim_{h \to 0} \frac{4+h - 4}{h (\sqrt{4+h} + 2)} = \lim_{h \to 0} \frac{h}{h (\sqrt{4+h} + 2)}
   \]
   
2. **Cancel out the \(h\):**
   \[
   \lim_{h \to 0} \frac{1}{\sqrt{4+h} + 2}
   \]
   
3. **Substitute \(h = 0\):**
   \[
   \frac{1}{\sqrt{4+0} + 2} = \frac{1}{2 + 2} = \frac{1}{4}
   \]

Thus, the solution for (a) is:
\[
\boxed{\frac{1}{4}}
\]

---

### (b) \(\lim_{x \to 4} \frac{x^2 - 16}{\sqrt{x} - 2}\)

1. **Factor the numerator:** 
   \[
   x^2 - 16 = (x-4)(x+4)
   \]
   So the limit becomes:
   \[
   \lim_{x \to 4} \frac{(x-4)(x+4)}{\sqrt{x} - 2}
   \]

2. **Multiply by the conjugate of the denominator:** To handle the \(\sqrt{x} - 2\) in the denominator, we multiply both the numerator and denominator by the conjugate, \(\sqrt{x} + 2\):
   \[
   \lim_{x \to 4} \frac{(x-4)(x+4)}{\sqrt{x} - 2} \times \frac{\sqrt{x} + 2}{\sqrt{x} + 2}
   \]
   Simplifying the denominator:
   \[
   \lim_{x \to 4} \frac{(x-4)(x+4)(\sqrt{x} + 2)}{(\sqrt{x})^2 - 2^2} = \lim_{x \to 4} \frac{(x-4)(x+4)(\sqrt{x} + 2)}{x - 4}
   \]

3. **Cancel out the \(x - 4\) terms:**
   \[
   \lim_{x \to 4} (x+4)(\sqrt{x} + 2)
   \]

4. **Substitute \(x = 4\):**
   \[
   (4+4)(\sqrt{4} + 2) = 8(2+2) = 8 \times 4 = 32
   \]

Thus, the solution for (b) is:
\[
\boxed{32}
\]

---

Let me know if you want further details or have any questions. Here are 5 related questions to extend this topic:

1. How do you find limits involving indeterminate forms?
2. What are some common techniques for simplifying radical expressions?
3. How do conjugates help in rationalizing denominators?
4. What is L'Hopital's Rule, and how does it apply to limits?
5. Can you explain the concept of continuity and its relationship to limits?

**Tip:** Multiplying by the conjugate is a powerful tool to simplify expressions involving square roots.

(a) lim h→0 √4+h−2 / h  (b) lim x→4 x²−16 / √x−2. Solve using algebraic simplification.

Learn how to solve limit problems involving radicals using algebraic simplification techniques. This problem explains the use of conjugates and factorization, showing step-by-step solutions to (a) lim h→0 √(4+h)−2 / h and (b) lim x→4 x²−16 / √x−2. Suitable for Grades 11-12.

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