Question

Selected values for the rate at which the radius of the second circle is changing

are provided below where v is a twice differentiable function and vis measured
in centimeters, t is measured in seconds and v(0) = 3cm.

t 0. 1. 2. 3. 4. 5. 6.
v'(t) 4. 2.1. 0. 1.2. -3 -2 1
Find
lim
t-0 (v(t)- 3e^t)/ (2v(t)-6)

Answer 1

The limit as
`\( t \)` approaches 0 for
`\((v(t) - 3e^t) / (2v(t) - 6)\) is \(1/8\),`determined using
`L'Hôpital's` Rule by taking the ratio of the derivatives of
`\(v(t)\)` with respect to \(t\).

To find the limit

`\[\lim_{{t \to 0}} \frac{{v(t) - 3e^t}}{{2v(t) - 6}}\]`

we can use
`L'Hôpital's Rule` , which states that if the limit of the ratio of two functions
`\( f(t)/g(t) \)` as
`\( t \)`approaches a certain value is an indeterminate form (e.g., \( 0/0 \) or
`\( \infty/\infty \)),` then the limit of the ratio is the same as the limit of the ratio of their derivatives
`\( f'(t)/g'(t) \).`

First, let's find the derivatives of
`\( v(t) \) with respect to \( t \):`

`\[ v'(t) = 4, \, 2.1, \, 0, \, 1.2, \, -3, \, -2, \, 1 \]`

Now, let's apply L'Hôpital's Rule to the given limit:

`\[\lim_{{t \to 0}} \frac{{v(t) - 3e^t}}{{2v(t) - 6}} = \lim_{{t \to 0}} \frac{{v'(t) - 3e^t}}{{2v'(t)}}\]`

Substitute the values of
`\( v'(t) \) at \( t = 0 \):`

`\[= \frac{{4 - 3}}{{2 \cdot 4}} = (1)/(8)\]`

So, the limit is
`\( (1)/(8) \).`