

 
LIMIT LAWS: Suppose that is a constant and the limits exist. Then 1. 2. 3. 4. 5. if 6. where is a positive integer 7. 8. 9. where is a positive integer 10. where is a positive integer (if is even, we assume that ) 11. where is a positive integer (if is even, we assume that ) EXAMPLES: Solution: We have Solution: We have DIRECT SUBSTITUTION PROPERTY: If is a polynomial or a rational function and is in the domain of then REMARK: The trigonometric functions also enjoy the Direct Substitution Property. EXAMPLES: 3. Solution (short): We have One can check that if then if then Both are very close to (which is ). This suggests that Compare with Note that in both cases we have indeterminate forms. The second limit is 1, since
In other words, the zero term gets canceled and the
limit perfectly exists as a finite number.
To apply the same idea to we first factor the bottom and then cancel the
zero term :
We first note that we can not use the Direct Substitution Property to evaluate ,
since the denominator is 0 at . In other words,
2 is not from the domain of ,
therefore DSP can not be used. However, this does
not necessarily mean that the limit does not exist. Actually, this limit does exist.
To understand it, we first note that the problem of finding the
limit is different from
the problem of evaluating at The latter is
impossible, since we can't divide by 0. Nevertheless, the limit of
when is goes to 2 does exist. Indeed, by
the nature of a limit when we are evaluating we are basically trying to see what
happens to when is close to 2 (as
close to 2 as we like), but not equal to 2 (see Definition of limit). One can check that
if then if then Both are very close to (which is ). This suggests that
However, our goal here is to prove it rigorously using the Limit Laws. To this end, we note that the
numerator of is also 0
when In other
words, when approaches 2, the top and the bottom of go to zero.
In this case we say that the limit is a indeterminate form. To evaluate this limit, we note that since (as we
discussed before), then, by algebra, we have
STEP A1: We show that . WORK: By arithmetic and (by definition of second power). Therefore hence STEP A2: We show that . WORK: Here we use the identity In our case and therefore hence STEP A3: We show that . WORK: Multiplication by 1 does not change the expression. Therefore hence STEP A4: We show that . WORK: We have In our case therefore So, we conclude that if therefore and now we can use the Direct Substitution Property to find . We have hence Solution (full): We first note that is type of an indeterminate form. Indeed, the top and the bottom of go to 0 when goes to 2: 4. Solution: We first note that is type of an indeterminate form. Indeed, the top and the bottom of go to 0 when goes to : and when . 5. Solution: We have 6. Solution: We first note that is type of an indeterminate form. Indeed, the top and the bottom of go to 0 when goes to 3: and when . which does not exist (division of nonzero by zero). 7. Solution (short): We have Solution (full): We first note that is type of an indeterminate form. Indeed, the top and the bottom of go to 0 when goes to 0: when . 8. Solution (short): We have Solution (full): We first note that is type of an indeterminate form. Indeed, the top and the bottom of go to 0 when goes to 0: when . Solution (full): We first note that is type of an indeterminate form. Indeed, the top and the bottom of go to 0 when goes to 0: when . 9. Solution (short): We have Solution (full): We first note that is type of an indeterminate form. Indeed, if is sufficiently close to 1 (on either side of 1) but not equal to 1, then and are large (or large negative). 10. Solution (short): We have Solution (full): We first note that is type of an indeterminate form. Indeed, the top and the bottom of go to 0 when goes to 2: when . 11. Solution 1 (short): We have Solution 1 (full): We first note that is type of an indeterminate form. Indeed, the top and the bottom of go to 0 when goes to 3: when . Solution 2 (short): We have Solution 2 (full): We first note that is type of an indeterminate form. Indeed, the top and the bottom of go to 0 when goes to 3: when . 12. Solution (short): We have Solution (full): We first note that is type of an indeterminate form. Indeed, the top and the bottom of go to 0 when goes to 2: when . 13. Solution 1 (short): We have Solution 1 (full): We first note that is type of an indeterminate form. Indeed, the top and the bottom of go to 0 when goes to 4: when . Solution 2 (short): We have Solution 2 (full): We first note that is type of an indeterminate form. Indeed, the top and the bottom of go to 0 when goes to 4: when . 14. Solution 1 (short): We have Solution 1 (full): We first note that is type of an indeterminate form. Indeed, the top and the bottom of go to 0 when goes to 0: when . Solution 2 (short): We have Solution 2 (full): We first note that is type of an indeterminate form. Indeed, the top and the bottom of go to 0 when goes to 0: when . 15. Solution (short): We have Solution (full): We first note that is type of an indeterminate form. Indeed, the top and the bottom of go to 0 when goes to 2: when . 16. Solution (short): We have Solution (full): We first note that is type of an indeterminate form. Indeed, the top and the bottom of go to 0 when goes to 0: when . 17. Solution (short): We have which is since Solution (full): We first note that is type of an indeterminate form. Indeed, the top and the bottom of go to 0 when goes to 0: when . which is since 18. Solution (short): We have Solution (full): We first note that is type of an indeterminate form. Indeed, if is sufficiently close to 0 (on either side of 0) but not equal to 0, then is sufficiently close to 0 and therefore is large (or large negative). 19. Solution (short): We have Solution (full): We first note that is type of an indeterminate form. Indeed, the top and the bottom of go to 0 when goes to 1: when . 20. Solution (short): We have Solution (full): We first note that is type of an indeterminate form. Indeed, the top and the bottom of go to 0 when goes to 0: when . We compute and separately. We have and therefore hence 21. Solution (short): We have Solution (full): We first note that is type of an indeterminate form. Indeed, the top and the bottom of go to 0 when goes to 0: when . 22. Let Find, if possible, and We have Since it follows that does not exist. 23. Let Find, if possible, We have (a) (b) (c) Since , it follows that does not exist. (d) (e) (f) (g) Since , it follows that . (h) 24. Show that does not exist. Solution (short): We have and Since it follows that does not exist. Solution (full): We have therefore Putting here , we get hence From this it follows that and Since it follows that does not exist. THE SQUEEZE THEOREM: If when is near (except possibly at ) and then . 25. Show that We first note that Multiplying all three parts of this inequality by we get Since it follows that by the Squeeze Theorem. 