Evaluate the limit

limh→05(4+h)2−80h=limh→05(4+h)2−80h=

*functions*

*equation editor*

**Hint:** Simplify the expression first, then take the limit.

We are given the limit:  

\[
\lim_{h \to 0} \frac{(4+h)^2 - 80}{h}
\]

### **Step 1: Expand the Numerator**  
Expanding \( (4+h)^2 \):

\[
(4+h)^2 = 16 + 8h + h^2
\]

Thus, the numerator becomes:

\[
(16 + 8h + h^2) - 80
\]

\[
16 + 8h + h^2 - 80 = h^2 + 8h - 64
\]

### **Step 2: Rewrite the Limit Expression**  

\[
\lim_{h \to 0} \frac{h^2 + 8h - 64}{h}
\]

### **Step 3: Factor and Simplify**  
We split the terms in the numerator:

\[
\frac{h^2 + 8h - 64}{h} = \frac{h(h + 8) - 64}{h}
\]

Splitting the fraction:

\[
\frac{h(h+8)}{h} - \frac{64}{h}
\]

The first term simplifies:

\[
h+8 - \frac{64}{h}
\]

### **Step 4: Evaluate the Limit**  
Taking the limit as \( h \to 0 \):

\[
\lim_{h \to 0} (h + 8 - \frac{64}{h})
\]

- \( h + 8 \to 8 \)
- \( \frac{64}{h} \) diverges to \( \pm \infty \)

Since \( \frac{64}{h} \) does not tend to a finite value, the limit **does not exist (DNE)**.  

### **Conclusion:**  
\[
\lim_{h \to 0} \frac{(4+h)^2 - 80}{h} \text{ does not exist.}
\]

Would you like further details or explanations?

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### **Related Questions:**
1. What happens if we approach the limit from the left (\( h \to 0^- \)) versus the right (\( h \to 0^+ \))?
2. How does this problem relate to the definition of a derivative?
3. What modifications would make this limit exist?
4. Can we use L'Hôpital’s Rule to evaluate this limit?
5. What is the graphical interpretation of this limit?

**Tip:** Always check for division by zero before evaluating limits.

Evaluate the limit limh→05(4+h)2−80h=limh→05(4+h)2−80h. Hint: Simplify the expression first, then take the limit.

Discover how to evaluate the limit limh→0 [(4+h)^2 - 80]/h with step-by-step guidance. This solution explores key concepts in limits and algebra, including polynomial expansion and limit laws. Suitable for high school students in Grades 10-12.

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