2.6 Asymptotes

Section 2.6 Notes Page 1 2.6 Limits Involving Infinity; Asymptotes of Graphs Sometimes limits don’t always go to a single number. Sometimes they may go to infinity or negative infinity. You do not need to graph every function. You can utilize tables. Let’s look at some examples: EXAMPLE: Use the table below for f( ) 5 and find: lim f (x), lim f (x), lim f (x) x 3 x3 x3 x3 x -2.5 -2.9 -2.99 -2.999 -3 -3.001 -3.01 -3.1 -3.5 f(x) 10 50 500 5000 undef. -5000 -500 -50 -10 a.) lim f (x) This means we need to look at what the f(x) value is approaching as I get close to 3 from the x3 positive side. This means we look x values to the right of -3, or values bigger than -3. This includes all x values that begin with a -2. As I get closer to -3 the value are getting larger. For example, when x is -2.999 the f(x) value is 5000. From this I can conclude that the values will be approaching positive infinity  which is our answer. b.) x3f( ) This means we need to look at what the f(x)value is approaching as I get close to 3 from the negative side. This means we look x values to the left of -3, or values smaller than -3. This includes all x values that begin with a -3. As I get closer to -3 the value are getting smaller. For example, when x is -3.001 the f(x) value is -5000. From this I can conclude that the values will be approaching negative infinity  which is our answer. c.) x3f (x) From parts a and b we can conc lude that this limit does not exist (d.n.e.) since it approaches two different values from the left and from the right. 2x EXAMPLE: Find li x1 1 x This one you do not need to draw a graph. We also can’t cancel anything and we can not plug in 1 one since the bottom will be zero. For this I suggest using a test value for x. As in the example above we want to choose a value that is close to 1. Since we are approaching 1 from the right we a value that is slightly larger than one. You can pick whatever value you want, but I will pick x = 1.0001. If we put this in for x, we get: 21.0001  30001 . This is a very large negative number. Our conclusion is that the limit is approaching 11.0001 negative infinity . EXAMPLE: Find lim x  2 1 x0 x Again we want to use a test value for x. Since we are approaching 0 from the left, we want to pick a value that is slightly less than 0. You can pick whatever value you want, but I will pick x = -0.0001: 2 1 (0.0001)  10000. This is a large positive number. Our conclusion is that the limit is 0.0001 approaching positive infinity  . Section 2.6 Notes Page 2 x2 EXAMPLE: Find li 2 x4 x 16 Do you need to use a test point for this one? No, because you do not get a zero in the denominator if you plug in 4 for x. Therefore we can just plug in 4 to get the answer: x2 4 2 16 1 li 2  2   . Notice that even though we are approaching 4 from the left, we can still x4 x  16 (4) 16 32 2 just put 4 into the expression. 2 EXAMPLE: Find lim 6x  x 1 x1 4x  4x 3 2 What about a test value for this one? Well even though ½ will cause the denominator to be zero we should first factor to see if we can cancel anything out first: lim (3x 1)(2x 1) We can cancel out the 2x + 1. x1 (2x 3)(2x 1) 2 3x 1 lim Now we can plug in the ½ without getting a zero in the denominator, so no test value is needed. x1 2x 3 2  1 3 5 3  1   1  lim 3x 1   2  2  2  5 x1 2x 3  1  3  4 8 2 2   3  2  2 EXAMPLE: Find lim  x  cosx 2 This one you do need a test value since we can’t cancel anything out and we will get a zero in the denominator if we plug in . It might be easier to first get the decimal . It is about 1.57 if we check this on a 2 2 calculator. We need to approach this decimal from the positive side, so we should check a number that is slightly above 1.57. Again you can pick any number, but I will use x = 1.6:  2  68.49. This is a large positive number, so our conclusion is that the limit is approaching + cos(1.6) Limits at Infinity The next set of problems deal with limits as x approaches infinity or negative infinity. Most of these type of problems will be given in fraction form. The main technique for these is to divide each term in the numerator and denominator by the highest power of x we see in the DENOMINATOR. Section 2.6 Notes Page 3 3x2 EXAMPLE: Find the limit: lix x  To solve this we will divide each term in the numerator and denominator by the highest power of x we see in the DENOMINATOR. In this problem the highest power of x in the denominator will be x. 3x 2  x x x x 4 Now we simplify.  x x 2 3 lim x Now we take the limit of each term separately: x 4 1 x 2 lim3 lim x x x 30 3x  2 4   3 So x  3. lim1lim 10 x 4 x x x 3 2x EXAMPLE: Find the limit: lim 3 x 3x 1 3 3 The highest power of x in the denominator xs, so we divide everything in the top and botby x . 3 2 x  x3 x3 x 3 sioplify. 3x  1 x3 x3 3 2 3  2 lim x x We will take the limit of each term individually. x 1 3 x3 3 2 lim 3  lim 2 xx xx  0  0 0 So lim 3 x  0. 1 3 0 x 3x  1 lim3 lim 3 x x x Section 2.6 Notes Page 4 3 2x 2 EXAMPLE: Find the limit: lim x 3x 1 The highest power of x in the denominator is x , so we div