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.