Question

1. using the limit definition of the derivative, find f’(2) for f(x) = \sqrt{x + 2}.

Explanation:

Step1: Recall the limit definition of the derivative

The limit definition of the derivative at a point \( x = a \) is \( f'(a)=\lim_{h
ightarrow0}\frac{f(a + h)-f(a)}{h} \). Here, \( a = 2 \) and \( f(x)=\sqrt{x + 2} \), so we first find \( f(2+h) \) and \( f(2) \).

\( f(2+h)=\sqrt{(2 + h)+2}=\sqrt{h + 4} \)

\( f(2)=\sqrt{2+2}=\sqrt{4} = 2 \)

So the difference quotient is \( \frac{f(2 + h)-f(2)}{h}=\frac{\sqrt{h + 4}-2}{h} \)

Step2: Rationalize the numerator

To simplify the limit, we multiply the numerator and denominator by the conjugate of the numerator, which is \( \sqrt{h + 4}+2 \).

\( \frac{\sqrt{h + 4}-2}{h}\times\frac{\sqrt{h + 4}+2}{\sqrt{h + 4}+2}=\frac{(\sqrt{h + 4})^2-2^2}{h(\sqrt{h + 4}+2)} \)

Using the difference of squares formula \( (a - b)(a + b)=a^2 - b^2 \), the numerator becomes \( (h + 4)-4=h \)

So the expression simplifies to \( \frac{h}{h(\sqrt{h + 4}+2)} \)

Step3: Cancel out the common factor

We can cancel out the \( h \) in the numerator and denominator (assuming \( h
eq0 \), which is valid since we are taking the limit as \( h
ightarrow0 \), not evaluating at \( h = 0 \)):

\( \frac{h}{h(\sqrt{h + 4}+2)}=\frac{1}{\sqrt{h + 4}+2} \)

Step4: Evaluate the limit as \( h

ightarrow0 \)
Now we take the limit as \( h
ightarrow0 \):

\( \lim_{h
ightarrow0}\frac{1}{\sqrt{h + 4}+2}=\frac{1}{\sqrt{0 + 4}+2}=\frac{1}{2 + 2}=\frac{1}{4} \)

Answer:

\( \frac{1}{4} \)