A Few Examples of Limit Proofs x (7x 4) = 10 SCRATCH WORK First, we need to find a way of relating x < δ and (7x 4) 10 < ɛ. We will use algebraic manipulation to get this relationship. Remember that the whole point of this manipulation is to find a δ in terms of ɛ so that if x < δ is true for some x then it forces (7x 4) 10 < ɛ to be true for that x. So we will start with the ɛ term and manipulate it until we find the delta term in there. So (7x 4) 10 < ɛ 7x 4 10 < ɛ 7x 14 < ɛ 7(x ) < ɛ 7 x < ɛ 7 x < ɛ Now 7 x < ɛ x < ɛ 7 Now with the above string of equivalences, we can see that if we let δ = ɛ 7, then we have found a δ that will fulfill the definition because if x < δ then x < ɛ 7 7 x < ɛ 7x 14 < ɛ 7x 4 10 < ɛ (7x 4) 10 < ɛ So we have that for any ɛ given to us if we let δ = ɛ 7 and x < δ then this will force (7x 4) 10 < ɛ thus showing that the definition is fulfilled. Now let s write the proof. Let ɛ > 0 be given. Let δ = ɛ 7. So if 0 < x < δ then x < ɛ 7 7 x < ɛ 7x 14 < ɛ 7x 4 10 < ɛ (7x 4) 10 < ɛ Therefore we have shown that lim x (7x 4) = 10. Done By proving the limit, all we have really done is show that the definition is fulfilled. Let s do a few more. This time I will leave out a lot of the work shown above. x 6 ( 3 x + 4) = 0
SCRATCH WORK Again we need a way to relate x 6 < δ and ( 3x + 4) 0 < ɛ Let s start with the ɛ term and see if we can manipulate it down to be something related to the δ term. ( 3 x + 4) 0 < ɛ 3 x + 4 < ɛ 3 (x 6) < ɛ 3 x 6 < ɛ 3ɛ 3 x 6 < ɛ x 6 < Therefore if we let δ = 3ɛ 3ɛ then if x 6 < δ then this forces x 6 < 3 x 6 < ɛ 3 x 6 < ɛ 3 (x 6) < ɛ 3 x + 4 < ɛ 3 x + 4 0 < ɛ ( 3x + 4) 0 < ɛ So for our choice of δ we can show that the definition is fulfilled. Just a quick note on this one. You will be tempted to let δ = 3ɛ which doesn t make sense because in the definition both δ and ɛ must be positive and if ɛ is positive then if we let δ = 3ɛ then δ would be negative - which is very bad. Now, let s do the proof. Let ɛ > 0 be given. Let δ = 3ɛ. So if 0 < x 6 < δ then x 6 < 3ɛ 3 x 6 < ɛ 3 x 6 < ɛ 3 (x 6) < ɛ 3 x + 4 < ɛ 3 x + 4 0 < ɛ ( 3x + 4) 0 < ɛ Therefore we have shown that lim ( x 6 3x + 4) = 0. Done Make sure that you understand every implication ( ) given in the above proof. Make sure you know why one statement leads to the next statement. Now we will do a proof where we need to restrict δ. x (x 3x + 1) = 1 SCRATCH WORK So we need a way of relating x < δ and (x 3x + 1) ( 1) < ɛ Let s start with the ɛ term. (x 3x + 1) ( 1) < ɛ x 3x + 1 + 1 < ɛ x 3x + < ɛ (x 1)(x ) < ɛ x 1 x < ɛ Now we run into a slight problem because we ve got the δ term just like before but it s not multiplied by a constant,
it s multiplied by some function of x. The problem is that we can t just divide by x 1 to get that x < ɛ x 1 = δ because δ cannot be a function of x, only of ɛ. So to fix this, we will restrict δ and by restricting δ we will actually be restricting x because we are only concerned with when x < δ. Let s choose some initial value to use as a restriction for δ. We will start with the value of 1. So let s start by saying δ < 1. If δ < 1 then x < δ < 1 x < 1 1 < x < 1 1 < x < 3. Now with this domain of values for x, what is the largest that x 1 can be? We get an upper bound of meaning that x 1 < if x is between 1 and 3. So let s look at where we left off with our ɛ term. We had gotten to x 1 x < ɛ. Now with our restriction on δ (i.e. δ < 1) this gives us a restriction on x (i.e. 1 < x < 3) which in turn gives us an upper bound for x 1 namely x 1 <. Therefore if x < ɛ then x 1 x < ɛ because x 1 x < x < ɛ. So if we can get that x < ɛ then certainly x 1 x < ɛ. So with x < ɛ then x < ɛ which we will set δ equal to. So we will let δ = ɛ. Now let s do the proof. Let ɛ > 0 be given. Let δ = min{ ɛ, 1}. Consider that if 0 < x < δ < 1 then 1 < x < 1 1 < x < 3 x 1 <. Now if 0 < x < δ x < ɛ x < ɛ x 1 x < x 1 < ɛ (Make sure you know why this is true.) x 1 x < ɛ (x 1)(x ) < ɛ x 3x+ < ɛ x 3x+1+1 < ɛ (x 3x+1) ( 1) < ɛ Therefore we have shown that lim x (x 3x + 1) = 1. Done A few tips and tricks. 95% of the time, you should start with your ɛ term and algebraically manipulate it until you get something of the form g(x) x c < ɛ where g(x) is some function of x. Once you find the function g(x) from the above hint, restrict δ (which forces a restriction on x) so that you can get an upper bound for g(x). Use this upper bound for g(x) to find δ. So if we algebraically get to g(x) x c < ɛ and g(x) < B for some number B then if we can make B x c < ɛ then this will force g(x) x c to be less than ɛ because g(x) x c < B x c < ɛ
I ve included a few more examples of some proofs. I haven t put the scratch work in. Make sure that you can do the scratch work to get the proof that I have written down. Prove lim (x + 3) = 1 x 1 Let ɛ > 0 be given. Let δ = ɛ. So 0 < x ( 1) < δ x+1 < ɛ x+1 < ɛ x+1 < ɛ (x+1) < ɛ x+ < ɛ (x+3) 1 < ɛ. Therefore we have shown that lim (x + 3) = 1. Done x 1 Prove lim ( 3x + 1) = 7 x Let ɛ > 0 be given. Let δ = ɛ 3. So 0 < x ( ) < δ x + < ɛ 3 3 x + < ɛ 3 x + < ɛ 3(x + ) < ɛ 3x 6 < ɛ ( 3x + 1) 7 < ɛ. Therefore we have shown that lim ( 3x + 1) = 7. Done x x 4 (x 3x + ) = 6 Let ɛ > 0 be given. Let δ = min{ ɛ 6, 1}. Consider that if 0 < x 4 < δ < 1 then 1 < x 4 < 1 3 < x < 5 x + 1 < 6. Now 0 < x 4 < δ x 4 < ɛ 6 6 x 4 < ɛ x + 1 x 4 < 6 x 4 < ɛ x + 1 x 4 < ɛ (x + 1)(x 4) < ɛ x 3x 4 < ɛ x 3x + 6 < ɛ (x 3x + ) 6 < ɛ
Therefore we have shown that lim x 4 (x 3x + ) = 6. Done x x 4 = 4 Let ɛ > 0 be given. Let δ = min{ ɛ 8, 1}. Consider that if 0 < x < δ < 1 then 1 < x < 1 1 < x < 3 8 x 4 < 8. Now 0 < x < δ x < ɛ 8 8 8 8 x < ɛ x < 8 x < ɛ x < ɛ x 4 x 4 8 x 8x 16 < ɛ < ɛ 8x 16 x 4 x 4 x 4 < ɛ + 16 x 4 < ɛ + 4(x 4) x 4 < ɛ x 4 + 4 < ɛ ( ) ( 4) x 4 < ɛ Therefore we have shown that lim = 4. Done x x 4