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as seems clear from the graph. To prove this we must show that given any  > 0, we can nd  > 0 such that j f x 1j <  whenever 0 < x 1 < . Now since x > 1, f x 1 and so the proof consists in the triviality that j1 1j <  whenever 0 < x 1 < . (c) As x ! 3 from the left, f x ! 1, i.e., lim f x 1. A proof can be formulated as in (b).
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Fig. 3-12
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(d) Since lim f x 6 lim f x , lim f x does not exist.
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3.13. Prove that lim x sin 1=x 0.
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We must show that given any  > 0, we can nd  > 0 such that jx sin 1=x 0j <  when 0 < jx 0j < . If 0 < jxj < , then jx sin 1=xj jxjj sin 1=xj @ jxj <  since j sin 1=xj @ 1 for all x 6 0. Making the choice  , we see that jx sin 1=xj <  when 0 < jxj < , completing the proof.
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3.14. Evaluate lim
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2 . e 1=x
As x ! 0 we suspect that 1=x increases inde nitely, e1=x increases inde nitely, e 1=x approaches 0, 1 e 1=x approaches 1; thus the required limit is 2. To prove this conjecture we must show that, given  > 0, we can nd  > 0 such that     2  2 <  when 0 < x <  1 e 1=x        2 2 2e 1=x  2 2     1 e 1=x 2  1 e 1=x  e1=x 1  
Since the function on the right is smaller than 1 for all x > 0, any  > 0 will work when e ! 1. If   2 e1=x 1 1 1=x 2 1 2 1 0 <  < 1, then 1=x > , e > 1, > ln 1 ; or 0 < x < . <  when 2   x  ln 2= 1 e 1
1 3.15. Explain exactly what is meant by the statement lim 1 and prove the validity of this x!1 x 1 4 statement.
The statement means that for each positive number M, we can nd a positive number  (depending on M in general) such that 1 >4 x 1 4 when 0 < jx 1j < 
1 1 1 To prove this note that or 0 < jx 1j < p . > M when 0 < x 1 4 < 4 M M x 1 4 p Choosing  1= 4 M , the required results follows.
3.16. Present a geometric proof that lim
sin  1. 
Construct a circle with center at O and radius OA OD 1, as in Fig. 3-13 below. Choose point B on OA extended and point C on OD so that lines BD and AC are perpendicular to OD. It is geometrically evident that
FUNCTIONS, LIMITS, AND CONTINUITY
[CHAP. 3
Area of triangle OAC < Area of sector OAD < Area of triangle OBD i.e., Dividing by 1 sin , 2  1 < sin  cos  sin  1 cos  < or <  cos  sin  1. As  ! 0, cos  ! 1 and it follows that lim !0  cos  <
1 2 sin  cos 
B A tan G sin G O cos G D
< 1  < 1 tan  2 2
Fig. 3-13
THEOREMS ON LIMITS 3.17. If lim f x exists, prove that it must be unique.
x!x0
We must show that if lim f x l1 and lim f x l2 , then l1 l2 . By hypothesis, given any  > 0 we can nd  > 0 such that j f x l1 j < =2 j f x l2 j < =2 Then by the absolute value property 2 on Page 3, jl1 l2 j jl1 f x f x l2 j @ jl1 f x j j f x l2 j < =2 =2  i.e., jl1 l2 j is less than any positive number  (however small) and so must be zero. Thus l1 l2 . when when 0 < jx x0 j <  0 < jx x0 j < 
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