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EVALUATING LIMITS OF EXPONENTIAL FUNCTIONS , First, we consider the natural exponential function f(r) = et, where e is called the Euler number, and has valuc 2.718281.... EXAMPLE 1: Evaluate the lim ex. Solution. We will construct the table of values for f(r) = et. We start by approaching the number 0 from the left or through the values less than but close to 0. -1 0.36787944117 -0.5 0.60653065971 -0.1 0.90483741803 -0.01 0.99004983374 -0.001 0.90900049983 -0.0001 0.990900049983 -0.00001 0.99999000005 Intuitively, from the table above, lim et = 1. Now we consider approaching 0 from its right or through values greater than but close to 0. I f(2) 2.71828182846 0.5 1.6487212707 0.1 1.10517091808 0.01 1.01005016708 0.001 1.00109050017 0.000 1.000100005 0.00001 1.00001000005 From the table, as the values of r get closer and closer to 0, the values of f(r) get closer I-+0+ and closer to 1. So. lim ed = 1. Combining the two one-sided limits allows us to conclude that lim ex = 1. 2-+0 We can use the graph of /(r) = e to determine its limit as z approaches 0. The figure below is the graph of S (I) = . y. Looking at Figure 1.1, as the values of r approach 0, either . . . . .. . . . . . . .. . . . . . . . . . ... ... ...'.... " . . . . . .4 - from the right or the left, the values of /(r) will get closer . . . . . .. .. . . . . . . . . . .... and closer to 1. We also have the following: . . . . . J .. . .. -.:. . . . . ..... (a) lime' = e = 2.718... . .. .. ........... . . .. (b) lime = e? = 7.389... . . . . . . . ; . . . . J. .... . . . . . ..;.. . .. .' 1-+2 1. .. . . . . . . . . . ..... .. . . (c) lim e' = e = 0.367... P. . . . .. . . 2 - - 1 . . . . . . . . . . .. ! . . . . .. ! ...... .. . .. . . . . . . . ...H.. . . . -3 -2 0 2 TTopic: EVALUATING LIMITS OF LOGARITHMIC FUNCTIONS Now, consider the natural logarithmic function f(I) = Inz. Recall that Ins = log. I. Morcover, it is the inverse of the natural exponential function y = cf. EXAMPLE 2: Evaluate lim Inr. I-+1 Solution. We will construct the table of values for f(I) = Inr. We first approach the number 1 from the left or through values less than but close to 1. 0.1 -2.30258509299 0.5 -0.69314718056 0.9 -0.10536051565 0.99 -0.01005033585 0.999 -0.00100050033 0.9999 -0.000100005 0.99999 -0.00001000005 Intuitively, lim Inr = 0. Now we consider approaching 1 from its right or through values I-1- greater than but close to 1. I .' f (I) 2 0.69314718056 1.5 0. 1051651081 1.1 0.0953101798 1.01 0.00995033085 1.001 0.00099950033 1.0001 0.000099995 1.00001 0.00000999995 Intuitively, lim Inz = 0. As the values of r get closer and closer to 1, the values of f(I) get closer and closer to 0. In symbols, lim In r = 0. I-+ We now consider the common logarithmic function f(I) = log10 7. Recall that f(I) = 10510 I = log I. EXAMPLE 3: Evaluate lim log I. 3-+1 Solution. We will construct the table of values for /(I) = logr. We first approach the number 1 from the left or through the values less than but close to 1. idea. ... - I f(I) 0.1 -1 0.5 -0.30102999566 0.9 -0.045757-49056 ... .:0. 0.99 -0:0043648054 0.999 -0.00043451177 0.9999 -0.0000-4343161 0.99999 -0.00000434296Now we consider approaching 1 from its right or through values greater than but close to 1. 2 0.30102999566 1.5 0.17609125905 1.1 0.04139268515 1.01 0.00-432137378 1.001 0.00043407747 1.0001 0.00004342727 1.00001 0.00000434292 As the values of I get closer and closer to 1, the values of f(r) get closer and closer to 0. In symbols. lim log r = 0. Consider now the graphs of both the natural and common logarithmic functions. We can use the following graphs to determine their limits as x approaches 1.. f(x) = Ing ((a) = log 2 3 4 7 The figure helps verify our observations that lim In I = 0 and lim logr = 0. Also, based 7-+1 on the figure, we have (a lim In1 = 1 (d) lim log r = log 3 = 0.47... I -+ 1-+3 (b) lim log r = 1 (e) lim Inr = -00 I-10 2-+0+ (c) lim Inr = In 3 = 1.09... (f) lim log r = -00 1+ 0+Topic: Limit of Trigonometric Functions EXAMPLE 4: Evaluate lim sin r. Solution. We will construct the table of values for f(x) = sinr. We first approach 0 from the left or through the values less than but close to 0. -0.$11-17098-18 -0.5 -0.479423538G -0.1 -0.09983341661 -0.01 -0.00999983333 -0.001 [SGGGGGG000 0- -0.0001 -0.00009999909 -0.00001 GGGGGGOOOO0 0- f (x) 1 0.8414709848 0.5 0.4794255386 0.1 0.0998334166-4 0.01 0.00999983333 0.001 0.00099999983 0.0001 0.00009999999 0.00001 0.00000999999 As the values of r get closer and closer to 1, the values of f(r) get closer and closer to 0. In symbols, lim sin r = 0. I-+0 We can also find lim sin by using the graph of the sine function. Consider the graph of f (x) = sin I. The graph validates our observation in Example 4 that Jim sing = 0. Also, using the graph, we have the following: (a) lim sin z = 1. "(c) lim sinr =-1. b) lim sin z = 0. (d) lim sinr = 0. I-+1 I - -xTopic: Special Limits (A) INTRODUCTION We will determine the limits of three special functions; namely. f(t) = sin t 1: 9 ( 1 ) = 1 - cost -. and h(() = These functions will be vital to the computation of the derivatives of the sine cosine. and natural exponential functions in Chapter 2. (B) LESSON PROPER THREE SPECIAL FUNCTIONS We start by evaluating the function f(() = sin t EXAMPLE 1: Evaluate lim sing 1-0 1 Solution. We will construct the table of values for f(() = sin t -. We first approach the number 0 from the left or through values less than but close to 0. 1 f ( ! ) -1 0.84147099848 -0.5 0.9588510772 -0.1 0.0983341665 -0.01 0.9999833334 -0.001 0.9999998333 -0.0001 0.99999999983 Now we consider approaching 0 from the right or through values greater than but close to 0. . . . . f(t) 1 0.8414709848 0.5 0.9588510772 0.1 0.9983341665 0.01 0.9999833334 0.001 0.9999998333 0.0001 0.9999999983sin t Since lim - and lim sin t 1-+0- 140+ - are both equal to 1, we conclude that sin f lim - = 1. 1-+0 The graph of f(t) = " below confirms that the y-values approach I as { approaches 0. -2 EXAMPLE 2: Evaluate lim I - cost Solution. We will construct the table of values for g(t) = 1 - cost ". We first approach the number 1 from the left or through the values less than but close to 0. g(t) -0.1596976941 -0.5 -0.2448318762 -0.1 -0.0109583 1722 -0.01 -0.0019909583 -0.001 -0.000-1999999 -0.0001 -0.000005 Now we consider approaching 0 from the right or through values greater than but close to 0. a(t ) 0.15969769 11 0.5 0.24-183-48762 0.1 0.0199583 1722 0.00 19999583 0.001 0.000 1999099 0.0001 0.000005Since lim 1 - cost. = 0 and lim 1 - cost = 0, we conclude that - cost i . lim =0. 1-+0 Below is the graph of g(t) = 1 - cost We see that the y-values approach 0 as t tends to 0. 1 - cost -0.5- We now consider the special function h(t) = EXAMPLE 3: Evaluate lim - 1 1 -+ 0 Solution. We will construct the table of values for h(t) = - e -1. We first approach the number 0 from the left or through the values less than but close to 0. .t . h(t) . .... -1 0.6321205588 -0.5 0.7869386806 -0.1 0.9516258196 -0.01 0.9950166251 -0.001 0.9995001666 -0.0001 0.9999500016 Now we consider approaching 0 from the right or through values greater than but close to 0. t h(t) 1.718281828 0.5 1:297442541 0.1 1.051709181 0.01 1.005016708 0.001 1.000500167 0.0001 1.000050002Since lim 1 1 =1 and lim = 1, we conclude that S-0- t. lim 1 = 1 . r->0 t. The graph of h(t) = e - 1 below confirms that lim h(t) = 1. t. 1-0 - li - 1I. Evaluate the following limits by constructing the table of values. 1. lim 3" 5. lim tant 2. lim 52 #6. lim cost Answer: -1 3. lim log z 4. lim cos 1 #7. lim sinx Answer: 0II. Given the graph below, evaluate the following limits: . .... .fee... Dosuedjune.. . . . ...... .................................... . ...................... 1. liz 67 2. lim b 3. lim b" 3 41.2III. Given the graph of the cosine function f(x) = cos r, evaluate the following limits: 1. lim cos z 2. lim cos I .3. lim cost