Hoạt động khám phá 1 trang 37 SGK Toán 11 Chân trời sáng tạo tập 2 - CTST

Phương pháp giải:

a) Thay vào công thức \(\frac{{s\left( t \right) - s\left( 5 \right)}}{{t - 5}}\).

b), c) Sử dụng các quy tắc tính giới hạn.

Lời giải chi tiết:

a)

\(\begin{array}{*{20}{l}}
\begin{array}{l}
\begin{array}{*{20}{l}}
{\left[ {5;5,1} \right]}
\end{array}:t = 5,1\\
 \Rightarrow \frac{{s\left( t \right) - s\left( 5 \right)}}{{t - 5}} = \frac{{4,9.5,{1^2} - 4,{{9.5}^2}}}{{5,1 - 5}} = 49,49
\end{array}\\
\begin{array}{l}
\begin{array}{*{20}{l}}
{\left[ {5;5,05} \right]}
\end{array}:t = 5,05\\
 \Rightarrow \frac{{s\left( t \right) - s\left( 5 \right)}}{{t - 5}} = \frac{{4,9.5,{{05}^2} - 4,{{9.5}^2}}}{{5,05 - 5}} = 49,245
\end{array}\\
\begin{array}{l}
\begin{array}{*{20}{l}}
{\left[ {5;5,01} \right]}
\end{array}:t = 5,01\\
 \Rightarrow \frac{{s\left( t \right) - s\left( 5 \right)}}{{t - 5}} = \frac{{4,9.5,{{01}^2} - 4,{{9.5}^2}}}{{5,01 - 5}} = 49,049
\end{array}\\
\begin{array}{l}
\begin{array}{*{20}{l}}
{\left[ {5;5,001} \right]}
\end{array}:t = 5,001\\
 \Rightarrow \frac{{s\left( t \right) - s\left( 5 \right)}}{{t - 5}} = \frac{{4,9.5,{{001}^2} - 4,{{9.5}^2}}}{{5,001 - 5}} = 49,0049
\end{array}\\
\begin{array}{l}
\begin{array}{*{20}{l}}
{\left[ {4,999;5} \right]}
\end{array}:t = 4,999\\
 \Rightarrow \frac{{s\left( t \right) - s\left( 5 \right)}}{{t - 5}} = \frac{{4,9.4,{{999}^2} - 4,{{9.5}^2}}}{{4,999 - 5}} = 48,9951
\end{array}\\
\begin{array}{l}
\begin{array}{*{20}{l}}
{\left[ {4,99;5} \right]}
\end{array}:t = 4,99\\
 \Rightarrow \frac{{s\left( t \right) - s\left( 5 \right)}}{{t - 5}} = \frac{{4,9.4,{{99}^2} - 4,{{9.5}^2}}}{{4,99 - 5}} = 48,951
\end{array}
\end{array}\)

Ta thấy: \(\frac{{s\left( t \right) - s\left( 5 \right)}}{{t - 5}}\) càng gần 49 khi \(t\) càng gần 5.

b)

\(\begin{array}{*{20}{l}}
\begin{array}{l}
\mathop {\lim }\limits_{t \to 5} \frac{{s\left( t \right) - s\left( 5 \right)}}{{t - 5}} = \mathop {\lim }\limits_{t \to 5} \frac{{4,9{t^2} - 4,{{9.5}^2}}}{{t - 5}}\\
 = \mathop {\lim }\limits_{t \to 5} \frac{{4,9\left( {{t^2} - {5^2}} \right)}}{{t - 5}} = \mathop {\lim }\limits_{t \to 5} \frac{{4,9\left( {t - 5} \right)\left( {t + 5} \right)}}{{t - 5}}
\end{array}\\
{ = \mathop {\lim }\limits_{t \to 5} 4,9\left( {t + 5} \right) = 4,9\left( {5 + 5} \right) = 49}
\end{array}\)

c)

\(\begin{array}{*{20}{l}}
\begin{array}{l}
\mathop {\lim }\limits_{t \to {t_0}} \frac{{s\left( t \right) - s\left( {{t_0}} \right)}}{{t - {t_0}}} = \mathop {\lim }\limits_{t \to 5} \frac{{4,9{t^2} - 4,9.t_0^2}}{{t - {t_0}}}\\
 = \mathop {\lim }\limits_{t \to 5} \frac{{4,9\left( {{t^2} - t_0^2} \right)}}{{t - t_0^2}} = \mathop {\lim }\limits_{t \to 5} \frac{{4,9\left( {t - {t_0}} \right)\left( {t + {t_0}} \right)}}{{t - {t_0}}}
\end{array}\\
{ = \mathop {\lim }\limits_{t \to 5} 4,9\left( {t + {t_0}} \right) = 4,9\left( {{t_0} + {t_0}} \right) = 9,8{t_0}}
\end{array}\)

-- Mod Toán 11 HỌC247