Using the definition of the derivative evaluate the following limits

1) Lim h---> 0

[ ( 8 + h )^1/3 - 2 ] / h

2) Lim x ---> pi/3

( 2cosx - 1 ) / ( 3x - pi)

Drag out L'Hospital's Rule

1) 1/3 (8+h)^(-2/3) / 1
= 1/3 * 1/4 = 1/12

2) -2sinx/3 = -√3/3

Oops. Using definition of derivative.

Check back later. Lots of messy algebra.

Use the binomial theorem to expand (8+h)1/3

(8+h)1/3 - 2 = 81/3 + 1/3 8-2/3h - 1/3 * 2/3 8-5/3h2 + ...
= -2 + 2 + 1/3 * 1/4 h - 2/9 * 1/32 h2 + ...

Divide by h and all the terms with h2 or higher go away, leaving only:

1/3 8-2/3 = 1/12

To evaluate the limits using the definition of the derivative, we need to find the derivative of the given expression and then substitute the limit value. Let's solve each case step by step.

1) Lim h ---> 0 [ ( 8 + h )^1/3 - 2 ] / h

To find the derivative, we'll use the power rule for differentiation:

f'(x) = n * x^(n-1)

In this case, we have ( 8 + h )^(1/3). Applying the power rule, we get:

f'(h) = 1/3 * ( 8 + h )^(-2/3)

Next, we need to substitute h with 0 to find the derivative at the limit:

f'(0) = 1/3 * ( 8 + 0 )^(-2/3) = 1/3 * 8^(-2/3) = 1/3 * (1/2) = 1/6

Therefore, the limit can be found by substituting the derivative:

Lim h ---> 0 [ ( 8 + h )^1/3 - 2 ] / h = f'(0) = 1/6

2) Lim x ---> pi/3 ( 2cosx - 1 ) / ( 3x - pi)

To find the derivative, we'll differentiate each term separately using the rules of differentiation.
For the first term, 2cosx, the derivative is -2sinx.
For the second term, 3x - pi, the derivative is 3.

So, the derivative of the given expression is:

f'(x) = -2sinx / 3

To find the limit at x approaching pi/3, we substitute x with pi/3 in the derivative:

f'(pi/3) = -2sin(pi/3) / 3 = -2(√3/2) / 3 = -√3/3

Therefore, the limit can be found by substituting the derivative:

Lim x ---> pi/3 ( 2cosx - 1 ) / ( 3x - pi) = f'(pi/3) = -√3/3

So, the limits using the definition of the derivative are:

1) Lim h ---> 0 [ ( 8 + h )^1/3 - 2 ] / h = 1/6

2) Lim x ---> pi/3 ( 2cosx - 1 ) / ( 3x - pi) = -√3/3