1.2 Evaluating Limits Algebraically Properties of Limits: ( )limx cf x� 75 3 21. lim : lim 7 lim 7 lim3 3x c x x xb b ex� � � �= = = = ( )7f x = x 4.9 4.99 4.999 4.9999 5.0001 5.001 5.01 5 f(x) 7 7 7 7 7 7 72. limx cx c�= ( )f x x= ex: x 1.9 1.99 1.999 2.001 2.01 f(x) 1.9 1.99 1.999 2.001 2.013. limn nx cx c�= (when n is a positive integer) Ex: ( ) ( )2 32 2 3 32 2lim 2 4 lim 2 8x xf x x f x xx x� �= == = = =4.Scalar Multiple Property ( ) ( )lim limx c x cb f x b f x� �� = � *The constant factor “b” can move through limits. This works for any constant “b” and any function.( )2 223 lim3 12xf x x x�= =2 22 2lim3 3 lim 3 4 12x xx x� �= � = �=5.Sum or Difference Property( ) ( ) ( ) ( )lim lim limx c x c x cf x g x f x g x� � �� = �� �� �ex:( ) ( ) ( ) ( )2 22 2lim limx xf x x g x x f x g x x x� �� �= = + = +� �� �� �2 2 22 2 22 lim lim lim 4 2 6x x xy x x x x x� � �� �= + + = + = + =� �6.Product Rule( ) ( ) ( ) ( )lim lim limx c x c x cf x g x f x g x� � �� = �� �� � *The limit of a product is the product of the limits.7.Quotient Rule( )( )( )( )limlimlimx cx cx cf xf xg x g x���� �=� �� � * The limit of a quotient is the quotient of the limits.8.( ) ( )lim limnnx c x cf x f x� �� �=� �� �� �Examples: 23lim 2 5xx x�� �- +� � Property (5) = 23 3 3lim lim 2 lim5x x xx x� � �- +Property (4) = 23 3 3lim 2lim lim 5x x xx x� � �- +Property (3) = 23 2 3 5- �+ =9 6 5 3 5 8- + = + =☼ You can use direct substitution to calculate the limit of a function IF the function is continuous at “c”.* In the example you will just substitute in the “c” value into the equation. ( ) ( )23lim 3 2 3 5 8x�� �- + =� �☼ A function is not continuous when there are holes at “c”. If you try direct substitution it will yield00 whenever there is a hole at “c” * This is most important to keep in mind when working with piecewise defined functions, where a break or jump in the function may exist.* For a piecewise defined function to have a limit the function’s pieces have to get the same number when you do direct substitution.☼ Also when you do direct substitution and get #0not zero, it is not the answer to the limit, but it does tell you where an asymptote is, and that a limit does not exist at that point for that function.☼ Direct substitution will work for any polynomial, as well as any trig function (where it is defined) and any log function (on its domain).To calculate limits analytically there are three main approaches.First try direct substitution, keeping in mind what your function looks like.If you get00, then try factor and cancel second. (If you get #0then the limit does not exist (DNE))**** Remember that piecewise functions are the exception to direct substitution; may be able to get an answer, but that answer may not be the answer to the limit.Example:225 6: lim2xx xFindx�- +- Direct substitution yields00. Now try to factor and cancle.( )2225 6lim23 2limxxX XXX X��- +=-- -=( )2X -( )( )2lim 3xX�= - Now use direct substitution! 2 3 1= - = -Factor and Cancel:3 226 4 241. lim2xx x xx�-- - ++ We can use synthetic division to help us factor here! 2 1 6 4 242 16 24- - -- -1 -8 12 022limxx�-+=( )( )28 122x xx- ++( ) ( ) ( )22 8 2 12 32= - - - + = * Canceling (x+2) also shows us that if we used direct substitution at the start, with (-2), we would have gotten00! Now that we have factored and canceled we can use direct substitution to find the answer to the limit.We could have also factored this problem from the start…it just takes a little more work. Compare for yourself:3 226 4 241. lim2xx x xx�-- - ++( ) ( )( )( )( )222226 4 6lim26 2lim26 2limxxxx x xxx xxx x�-�-�-- - -=+- -=+- +=( )( )22xx-+Now we will use direct substitution ( ) ( ) ( ) ( ) ( ) ( )2lim 6 2 2 6 2 2 8 4 32xx x�-= - - = - - - - = - - =with a hole at (-2,32)Either way you choose to do this problem is correct; it just might not always be obvious how to factor a problem such as this. In the end, for either method the answer is still 32.**** Don’t forget to write down 2limx�-=all the way through your work, until you do direct substitution!4 2333 1052. lim4 15xx x xx x�-+ + -- + * First try direct substitution! Here we get00, so we will need to try something else. We will try factoring with the help of synthetic division next.Using our “c” value will help us with factoring.3 1 0 3 1 1053 9 36 1051 3 12 35 0- -- -- -00 3 1 0 4 153 5 0- -- *Getting 0s helps us find our indetermanent!*If we would have gotten 050=Limit answer would be 0!*If we would have gotten 50=Does Not Exist (DNE) (asymptote here instead) Now we will see what happens when we factor and cancel:33limxx�-+=( )( )3 23 12 353x x xx- + -+( )( )23 5x x- + Try direct substitution again…-3 is no longer in our list of p/q so 0/0 will not happen again.3 1 3 12 353 18 901 6 30 125- -- -- - 3 1 3 53 181 6 23- ---12523-=************************************************************************************************************The third method for finding limits is conjugation!2553. lim25xxx�-- (D.S. yields00!)25 2555255lim lim25x xxx xxx� �� �+�--=�� �+�� =-�( )25x -( )( ) ( )251 1 1 1lim5 5 1025 55 5xx x�= = = =+++ +* You cannot use synthetic division, like in the previous examples, with radicals; although, this problem could have also been factored from the start. 2553. lim25xxx�--( )( )2225 2555lim lim5x xxxx� �--= =-( )5 5x x+ -()251 1 1 1lim5 5 105 25 5xx�= = = =++ +FIND:01 14. limxxx�+ - (D.S. yields00!)( )0 0 01 1 1 1lim l1 11 1im lim1 1x x xxx x xxx xx� � �+ - + -=+ ++ +� = =+ + x( )01lim1 11 1xxx�= =+ ++ +1 1 1 11 1 20 1 1 1 1= = = =++ + …