Basic Calculus Learning Activity Sheet PDF

# Learning Activity Sheet (LAS) No. 1 ## BASIC CALCULUS LEARNING ACTIVITY SHEET **Illustrating limits involving the expressions $ \frac{sin(t)}{t} $ , $ \frac{1-cos(t)}{t} $ and using tables of values** ### Learning Competency with Code Illustrates limits involving the expressions $ \frac{sin(...

# Learning Activity Sheet (LAS) No. 1 ## BASIC CALCULUS LEARNING ACTIVITY SHEET **Illustrating limits involving the expressions $ \frac{sin(t)}{t} $ , $ \frac{1-cos(t)}{t} $ and using tables of values** ### Learning Competency with Code Illustrates limits involving the expressions $ \frac{sin(t)}{t} $ , $ \frac{1-cos(t)}{t} $ and using tables of values (STEM_BC11LC-IIIb-2) ### Background Information for Learners #### Special Limits of Transcendental Functions This learning activity focuses on determining the limits of three special functions. These are $ f(t) = \frac{sin(t)}{t} $ , $ f(t) = \frac{1-cos(t)}{t} $, $ h(t) = \frac{e^t - 1}{t} $ ### Accompanying DepEd Textbook and Educational Sites Basic Calculus Learner's Material. First Edition (2016). Pasig City: Department of Education-Bureau of Learning Resources (DepEd-BLR), pp 62-72. Leithold, L. 2002. The Calculus 7. Pearson Education Asia Pte Ltd 23-25 First Lok Yang Road Singapore 629733. ### Activity Proper #### Example 6: Evaluate the $lim_{t \to 0} \frac{sin(t)}{t} $ **(Note: The values on the given table below were computed using scientific calculator set to radian mode.)** **Solution:** Let us construct the table of values for $g(t) = \frac{1-cos(t)}{t}$ starting with values of t approaching 0 from the left or through values less than and infinitesimally close to 0. Similarly, we construct the table of values for g(t) starting with the values of t approaching 0 from the right or through the values greater than and infinitesimally close to 0. | t | $ g(t) = \frac{1 - cos(t)}{t} $ | |---|---| | -1 | -0.459697694 | | -0.5 | -0.244834876 | | -0.1 | -0.049958347 | | -0.01 | -0.0049999583 | | -0.001 | -0.0005 | | -0.0001 | -0.00004999 | |0.0001 | 0.00004999 | | 0.001 | 0.0005 | | 0.01 | 0.0049999583 | | 0.1 | 0.049958347 | | 0.5 | 0.244834876 | | 1 | 0.459697694 | **Remarks:** - Based from table 6 and graph on figure 5, as the values of t get closer and closer to 0 from the left, the values of $ f(t) = \frac{sin(t)}{t} $ get closer and closer to 1. Hence, $ lim_{t \to 0^{-}} \frac{sin(t)}{t} = 1$. - Similarly, as values of t approaches 0 from the right, the values of $ f(t) = \frac{sin(t)}{t} $ approaches 1. Thus, $lim_{t \to 0^{+}} \frac{sin(t)}{t} = 1$ - Since the two-sided limits exist and they have a common value of 1, then $ lim_{t \to 0} \frac{sin(t)}{t} = 1$ #### Example 8: Evaluate the $lim_{t \to 0} \frac{e^t - 1}{t}$ **Solution:** Let us construct table 8 of values for $h(t) = \frac{e^t - 1}{t} $ , we start with values of t approaching 0 from the left or through values less than and infinitesimally close to 0. Similarly, we construct the table of values for h(t) starting with the values of t approaching 0 from the right or through the values greater than and infinitesimally close to 0. | t | $h(t) = \frac{e^t - 1}{t} $ | |---|---| | -2 | 0.632120558 | | -1.5 | 0.78693868 | | -1.1 | 0.951625819 | | -1.01 | 0.995016624 | | -1.001 | 0.999500165 | | -1 | undefined | | -0.999 | 1.00050017 | | -0.99 | 1.005016708 | Since the two-sided limits exist and they have a common value 1, then $lim_{t \to -1} \frac{e^{t+1} - 1}{t + 1} = 1 $. **Remarks:** Based on the above discussion, we have the following results. - $lim_{t \to 0} \frac{sin(t)}{t} = 1$ - $ lim_{t \to 0} \frac{1 - cos(t)}{t} = 0$ - $lim_{t \to 1} \frac{e^{t + 1} - 1}{t+1} = 1$ #### Example 9: Evaluate the $lim_{t \to -1} \frac{sin(t + 1)}{t + 1} $ **(Note: The values on the given table below were computed using scientific calculator set to radian mode.)** **Solution:** **Method 1:** Let us construct the table of values for $g(t) = \frac{1-cos(t-1)}{t - 1} $ , let x=t-1, if t = 1, then x = 0. Hence, $lim_{t \to 1} \frac{1-cos(t - 1)}{t - 1} = lim_{x \to 0} \frac{1-cos(x)}{x} = 0$. **Method 2:** We start with values of t approaching -1 from the left or through values less than and infinitesimally close to -1. Similarly, we construct the table of values starting with the values of t approaching -1 from the right or through the values greater than and infinitesimally close to -1. Refer to table 11 below. | t | $h(t) = \frac{e^{t + 1} - 1}{t + 1} $ | |---|---| | -2 | 0.632120558 | | -1.5 | 0.78693868 | | -1.1 | 0.951625819 | | -1.01 | 0.995016624 | | -1.001 | 0.999500165 | | -1 | undefined | | -0.999 | 1.00050017 | | -0.99 | 1.005016708 | Hence, $lim_{t \to -1} \frac{sin(t + 1)}{t + 1} = 1$. #### Example 10: Evaluate the $lim_{t \to 1} \frac{e^{t + 1} - 1}{t + 1} $ **(Note: The values on the given table below were computed using scientific calculator set to radian mode.)** **Solution:** **Method 1** Let us construct the table of values for g(t) = lim t→1 $ \frac{1 - cos(t - 1)}{t - 1}$. Let x=t-1, ift = 1, then x = 0. Hence, lim $t \to 1$ $\frac{1 - cos(t - 1)}{t - 1} = lim_{x \to 0} \frac{1 - cos(x)}{x} = 0$. **Method 2:** We start with values of t approaching -1 from the left or through values less than and infinitesimally close to -1. Similarly, we construct the table of values starting with the values of t approaching -1 from the right or through the values greater than and infinitesimally close to -1. Refer to table 11 below. | t | $h(t) = \frac{e^{t + 1} - 1}{t + 1} $ | |---|---| | -2 | 0.632120558 | | -1.5 | 0.78693868 | | -1.1 | 0.951625819 | | -1.01 | 0.995016624 | | -1.001 | 0.999500165 | | -1 | undefined | | -0.999 | 1.00050017 | | -0.99 | 1.005016708 |