Question

1. (limlimits_{h \to 0} \frac{sqrt{16 + h} - 4}{h})

Explanation:

Step1: Rationalize the numerator

Multiply the numerator and denominator by the conjugate of the numerator, which is $\sqrt{16 + h} + 4$.
$$\lim_{h \to 0} \frac{\sqrt{16 + h} - 4}{h} \cdot \frac{\sqrt{16 + h} + 4}{\sqrt{16 + h} + 4}$$

Step2: Simplify the numerator

Using the difference of squares formula $(a - b)(a + b) = a^2 - b^2$, the numerator becomes $(\sqrt{16 + h})^2 - 4^2 = 16 + h - 16 = h$.
So the expression simplifies to:
$$\lim_{h \to 0} \frac{h}{h(\sqrt{16 + h} + 4)}$$

Step3: Cancel out h

Cancel the common factor $h$ in the numerator and denominator (since $h
eq 0$ when taking the limit as $h \to 0$).
$$\lim_{h \to 0} \frac{1}{\sqrt{16 + h} + 4}$$

Step4: Substitute h = 0

Substitute $h = 0$ into the simplified expression.
$$\frac{1}{\sqrt{16 + 0} + 4} = \frac{1}{4 + 4} = \frac{1}{8}$$

Answer:

$\frac{1}{8}$