The processor speeds, in megahettz (MHz), of Intel processors during the period 1996-2010 can be approximated by the following function of time \( t \) in years since the start of 1990 : \[ v(t)=\left\{\begin{array}{cc} 400 t-2200 & \text { if } 6 \leq t<15 \\ 3900 & \text { if } 15 \leq t \leq 20 \end{array}\right. \] a. Compute \( \lim _{t \rightarrow 8} v(t) \). 8 falls in the equation \( 6 \leq t<15 \). So, we will use \( 400 t-2200 \). \[ \begin{array}{l} =400(8)-200 \\ =1,000 . \quad \text { Therefore, } \lim _{t \rightarrow 8} v(t)=1,000 . \end{array} \] b. Compute \( \lim _{t \rightarrow 15^{-}} v(t) \) and \( \lim _{t \rightarrow 15^{+}} v(t) \). c. Compute \( \lim _{t \rightarrow 15} v(t) \)

To compute \( \lim _{t \rightarrow 15^{-}} v(t) \): Since \( t = 15 \) is approaching from the left, we use the first case of the function, which is valid for \( 6 \leq t < 15 \): \[ \lim _{t \rightarrow 15^{-}} v(t) = 400(15) - 2200 = 6000 - 2200 = 3800. \] Next, for \( \lim _{t \rightarrow 15^{+}} v(t) \): Here, since \( t = 15 \) is approaching from the right, we will use the second case, which is valid for \( 15 \leq t \leq 20 \): \[ \lim _{t \rightarrow 15^{+}} v(t) = 3900. \] Finally, for \( \lim _{t \rightarrow 15} v(t) \): This limit checks for continuity at \( t = 15 \): \[ \lim _{t \rightarrow 15} v(t) = \lim _{t \rightarrow 15^{-}} v(t) = \lim _{t \rightarrow 15^{+}} v(t). \] Since: \[ \lim _{t \rightarrow 15^{-}} v(t) = 3800 \quad \text{and} \quad \lim _{t \rightarrow 15^{+}} v(t) = 3900, \] thus, \( \lim _{t \rightarrow 15} v(t) \) does not exist because the left-hand limit does not equal the right-hand limit.