Question

1. use the limit definition of the derivative to find the tangent to ( y = x^3 ) at ( x = 2 ).

Explanation:

Step1: Recall the limit definition of the derivative

The limit definition of the derivative of a function \( y = f(x) \) at a point \( x = a \) is \( f'(a)=\lim_{h
ightarrow0}\frac{f(a + h)-f(a)}{h} \). For the tangent line, we first need to find the derivative at \( x = 2 \) (which will be the slope \( m \) of the tangent line) and then use the point - slope form \( y - y_0=m(x - x_0) \), where \( (x_0,y_0) \) is the point of tangency.

First, find \( f(2) \) for \( f(x)=x^{3} \). Substitute \( x = 2 \) into \( f(x) \): \( f(2)=2^{3}=8 \). So the point of tangency is \( (2,8) \).

Step2: Compute the derivative at \( x = 2 \) using the limit definition

Let \( f(x)=x^{3} \), then \( f(2 + h)=(2 + h)^{3} \). Expand \( (2 + h)^{3} \) using the formula \( (a + b)^{3}=a^{3}+3a^{2}b + 3ab^{2}+b^{3} \), where \( a = 2 \) and \( b = h \). So \( (2+h)^{3}=2^{3}+3\times2^{2}\times h+3\times2\times h^{2}+h^{3}=8 + 12h+6h^{2}+h^{3} \).

Now, substitute into the limit definition of the derivative:
\[

$$\begin{align*} f'(2)&=\lim_{h ightarrow0}\frac{f(2 + h)-f(2)}{h}\\ &=\lim_{h ightarrow0}\frac{(8 + 12h+6h^{2}+h^{3})-8}{h}\\ &=\lim_{h ightarrow0}\frac{12h+6h^{2}+h^{3}}{h}\\ &=\lim_{h ightarrow0}\frac{h(12 + 6h+h^{2})}{h}\\ \end{align*}$$

\]
Since \( h
eq0 \) (we are taking the limit as \( h
ightarrow0 \), not evaluating at \( h = 0 \)), we can cancel out the \( h \) terms:
\[
f'(2)=\lim_{h
ightarrow0}(12 + 6h+h^{2})
\]
Now, take the limit as \( h
ightarrow0 \). Substitute \( h = 0 \) into \( 12 + 6h+h^{2} \): \( 12+6\times0 + 0^{2}=12 \). So the slope of the tangent line \( m = 12 \).

Step3: Find the equation of the tangent line

We use the point - slope form \( y - y_0=m(x - x_0) \), where \( x_0 = 2 \), \( y_0 = 8 \), and \( m = 12 \).
Substitute these values into the formula: \( y - 8=12(x - 2) \).
Expand the right - hand side: \( y-8 = 12x-24 \).
Add 8 to both sides to get the slope - intercept form: \( y=12x-16 \).

Answer:

The equation of the tangent line to \( y = x^{3} \) at \( x = 2 \) is \( y = 12x-16 \)