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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 \( \lim _{t \rightarrow 15} v(t) \)

Ask by Moreno Klein. in the United States
Mar 13,2025

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Answer

a. \( \lim_{t \to 8} v(t) = 1000 \) b. \( \lim_{t \to 15^-} v(t) = 3800 \) and \( \lim_{t \to 15^+} v(t) = 3900 \) c. \( \lim_{t \to 15} v(t) \) does not exist

Solution

a. For \( t = 8 \), we use the first equation (since \( 6 \leq t < 15 \)): \[ v(t)=400t-2200. \] Substitute \( t=8 \): \[ v(8)=400(8)-2200=3200-2200=1000. \] Thus, \[ \lim_{t \to 8} v(t)=1000. \] b. To compute the one-sided limits at \( t=15 \): - For the left-hand limit \( \lim_{t \to 15^-} v(t) \), we use the equation for \( 6\leq t<15 \): \[ v(t)=400t-2200. \] Substitute \( t=15 \): \[ \lim_{t \to 15^-} v(t)=400(15)-2200=6000-2200=3800. \] - For the right-hand limit \( \lim_{t \to 15^+} v(t) \), we use the equation for \( 15\leq t\leq 20 \): \[ v(t)=3900. \] Since the value is constant, \[ \lim_{t \to 15^+} v(t)=3900. \] c. Since the left-hand limit and the right-hand limit at \( t = 15 \) are different, \[ \lim_{t \to 15^-} v(t)=3800 \quad \text{and} \quad \lim_{t \to 15^+} v(t)=3900, \] the overall limit does not exist. That is, \[ \lim_{t \to 15} v(t) \quad \text{does not exist}. \]

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The Deep Dive

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.

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