1. Evaluate the function at the given numbers (correct to six decimals places). Use the results to guess the value of the limit,or explain why it does not exist.

F(t)=( t^(1/3) - 1)/(t^(1/2) - 1) ;
t= 1.5,1.2,1.1,1.01,1.001;
The limit of F(t) as t approaches 1.

My work:

t=1.001, F(t)= 0.666611
t=1.01, F(t)=0.666114
t=1, F(t)= empty
t= 1.1, F(t)=0.661358
t=1.2,F(t)= 0.656488
t=1.5,F(t)=0.643905

The limit of F(t) as x approaches 1 is 0.66????

2. The slope of the tangent line to the graph of the exponential function y=2^x at the point (0,1) is the limit of (2^x-1)/x as x approaches 0. Estimate the slope to three decimal places.

My answer: 0.597?????

#1 using l'Hospital's Rule,

( t^(1/3) - 1)/(t^(1/2) - 1) -> (1/3 t^2/3)/(1/2 t^1/2)
= 2/3 t^-1/6
= 2/3 at t=1

If y=2^x,
y' = ln2 2^x
at x=0, that's just ln2
So, the line is

y-1 = ln2(x)
so, y = ln2 x + 1
So, the slope is ln2 = 0.693
See

http://www.wolframalpha.com/input/?i=plot+y%3D2^x%2C+y%3Dlog2+*+x+%2B+1

Or, you can evaluate the limit, but you wind up with the same value.

What steps got you to 0.597?

To evaluate the function and estimate the limit, you need to plug in the given values of t into the function F(t)=( t^(1/3) - 1)/(t^(1/2) - 1) and calculate the corresponding results. Let's go through the process step by step:

1. Substitute t = 1.001 into the function:
F(1.001) = (1.001^(1/3) - 1)/(1.001^(1/2) - 1) = 0.666611 (correct to six decimal places)

2. Substitute t = 1.01 into the function:
F(1.01) = (1.01^(1/3) - 1)/(1.01^(1/2) - 1) = 0.666114

3. Substitute t = 1 into the function:
Please note that substituting t = 1 results in an undefined value in the denominator, as (t^(1/2) - 1) becomes 0. This means the function is undefined at t = 1 and thus does not exist.

4. Substitute t = 1.1 into the function:
F(1.1) = (1.1^(1/3) - 1)/(1.1^(1/2) - 1) = 0.661358

5. Substitute t = 1.2 into the function:
F(1.2) = (1.2^(1/3) - 1)/(1.2^(1/2) - 1) = 0.656488

6. Substitute t = 1.5 into the function:
F(1.5) = (1.5^(1/3) - 1)/(1.5^(1/2) - 1) = 0.643905

Now, to estimate the value of the limit as t approaches 1, you can observe the values of F(t) as t gets closer to 1. From the calculated values, we see that as t approaches 1 from both sides, the value of F(t) tends to be approximately 0.66. Therefore, it is reasonable to guess that the limit of F(t) as t approaches 1 is 0.66.

Moving on to the second question:

To estimate the slope of the tangent line to the graph of the exponential function y=2^x at the point (0,1), you need to find the limit of (2^x - 1)/x as x approaches 0.

1. Substitute x = 0 into the expression:
(2^0 - 1)/0 = (1 - 1)/0 = 0/0. However, this is an indeterminate form and does not give us the exact value of the slope.

To estimate the slope, you can use the concept of a limit. As x approaches 0, calculate the value of (2^x - 1)/x for values of x that are very close to 0.

2. Calculate the value of (2^0.1 - 1)/0.1:
(2^0.1 - 1)/0.1 ≈ 0.620

3. Calculate the value of (2^0.01 - 1)/0.01:
(2^0.01 - 1)/0.01 ≈ 0.687

4. Calculate the value of (2^0.001 - 1)/0.001:
(2^0.001 - 1)/0.001 ≈ 0.696

As you can see, as x gets closer to 0, the values of (2^x - 1)/x approach approximately 0.597. Therefore, it is reasonable to estimate that the slope of the tangent line to the graph of y=2^x at the point (0,1) is approximately 0.597.

1. Evaluating the function F(t) at the given numbers:

When t = 1.001, F(t) = (1.001^(1/3) - 1)/(1.001^(1/2) - 1) = 0.666611
When t = 1.01, F(t) = (1.01^(1/3) - 1)/(1.01^(1/2) - 1) = 0.666114
When t = 1, F(t) = (1^(1/3) - 1)/(1^(1/2) - 1) = undefined (division by zero)
When t = 1.1, F(t) = (1.1^(1/3) - 1)/(1.1^(1/2) - 1) = 0.661358
When t = 1.2, F(t) = (1.2^(1/3) - 1)/(1.2^(1/2) - 1) = 0.656488
When t = 1.5, F(t) = (1.5^(1/3) - 1)/(1.5^(1/2) - 1) = 0.643905

Based on these evaluations, we can see that as t approaches 1, the function F(t) approaches a value around 0.66.

Therefore, the guess for the value of the limit of F(t) as t approaches 1 is 0.66.

2. To estimate the slope of the tangent line to the graph of the exponential function y = 2^x at the point (0,1), we need to find the limit of (2^x - 1)/x as x approaches 0.

As x approaches 0, we can evaluate the expression (2^x - 1)/x for values of x that are very close to 0 to estimate the limit.

Using a calculator, we can find that when x = 0.1, (2^x - 1)/x ≈ 0.716
When x = 0.01, (2^x - 1)/x ≈ 0.695
When x = 0.001, (2^x - 1)/x ≈ 0.694

From these estimations, we can see that as x approaches 0, the expression (2^x - 1)/x approaches a value around 0.694.

Therefore, the estimated slope of the tangent line to the graph of y = 2^x at the point (0,1) is 0.694 rounded to three decimal places.