sin(x) < x sin(x) x < tan(x) sin(x) x cos(x) 1 < sin(x) sin(x) 1 < 1 cos(x) 1 cos(x) = 1 cos2 (x) 1 + cos(x) = sin2 (x) 1 < x 2

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1 . Problem Show that using an ɛ δ proof. sin() lim = 0 Solution: One can see that the following inequalities are true for values close to zero, both positive and negative. This in turn implies that On the interval ( π/, π/), this implies sin() < < tan() cos() < sin() < Subtracting from both sides, we have Taking absolute values, again, we have cos() < sin() cos() < sin() < < 0 sin() < cos() This step is important, since we can show that cos() goes to zero as does, that is, the right hand side can be found in terms of δ. Since cos() = cos () + cos() = sin () + cos() Putting this together we have Therefore, if 0 < δ then sin() < f() L = sin() δ Summing up, if δ = ɛ then f() L = sin() δ = ɛ

2 . Problem Using the results of the previous problem, show that eists. Solution: The easiest way is to write the problem as Let u = sin(), then we have sin(sin()) lim 0 sin(sin()) sin(sin()) sin() lim = lim lim 0 0 sin() 0 sin(sin()) sin(u) sin() lim = lim lim = ()() = 0 u 0 u 0

3 3. Problem 3 Show that does not eists, using an ɛ δ proof. lim sin(/) 0 Solution: The easiest way is a proof by contradiction. Suppose the limit did eist, then there would be an L such that given an ɛ > 0, then < δ would imply sin(/) L < ɛ. Choose an ɛ > 0. Find the δ, depending on ɛ. We can find an -value, e.g. = /(Nπ) such that < δ, so that sin(/ ) L = sin(nπ) L < 0 L < ɛ Similarly, we can find an -value, e.g. = /((N + /)π) so that < δ, so that sin(/ ) L = sin((n + /)π) L = L < ɛ This is a problem, since if ɛ < /, L can t be close to both 0 and! Intuitively, sin(/) oscillates to rapidly near = 0. It takes on values near -, 0, +, arbitrarily close to = 0 so it cannot approach a limit...

4 4. Problem 4 Show that Note: To show that lim [ + ] = 0 lim f() = L we must show that given any ɛ > 0, we can find an N, depending on ɛ, such that > N = f() L < ɛ The first step is to multiply by the conjugate lim [ + ] = lim [ ][ ] = lim + + = lim + + The critical observation is that this can be estimated in terms of N. < < + + N So if > N > 4/ɛ, then < < < ɛ + + N This gives the relationship between N and ɛ eplicitly.

5 5. Problem 5 The Fibonacci numbers are defined by the relationship with a 0 =, a =. What we want to show is that a(n) = a n = a n+ a n + a n, n =... ( + 5 ) n for all integer values of n. This is an indication that we must use induction. First we show that it is true for n =. a() = ( + 5 ) which is true. Now asssume that is true for n k, that is Then we eamine a(k) = a(k ) = and by combining common terms, = The crucial observation is that ( + 5 ( + 5 ( + 5 ( 5 ) n ( 5 ) 0 = = ) k ) k a(k + ) = a(k) + a(k ) = ( + 5 ( + 5 ) k ( ) k ) k ) ( 5 ) k ( 5 ) k ( 5 ) k ( 5 ) k ( 5 ) k ( + 5 ) so Therefore, ( + 5 ) = ( + 5 a(k + ) = ) k+ = ( + 5 ( + 5 = + ( + 5 ) ) k + ( + 5 ) k ) k ( 5 ) k+

6 To get the asymptotics, note that and since hence Therefore c = 5+ 5 and r = + 5. a n 5+ 5 ( + 5 ) n = ( )n < n 0 a n 5+ 5 ( + 5 ) n

7 6. Problem 6 Show that one can compute π by means of the infinite series π = 4 ( ) = ( ) n n + n=0 Solution: The function tan () can be written as tan () = + d = ( ) Substituting =, we have = π/4 = tan =

8 7. Problem 7 Show that the non-alternating series n=0 n + diverges. One can do this several ways. The direct approach is to group terms. If we take k successive terms, staring with N+ we have (taking k of the smallest terms as a lower bound) This will be greater than a fied constant, e.g. /4, if N + + N N N + k k N + k k N + k 4 that is, if k N /. So, letting k = N we have the inequality N + + N N N + k = N + + N N N + N N 4N 4 Since we have an infinite number of these groups, all of which are greater than /4, the series diverges. Another way of demonstrating this is to compare the series to an integral. Using the left hand rule for Riemanns sums Left Hand Endpoint y Riemann Sum f()=/ / /3 / This implies that = n=0 n + n = n= = [ ] [ N ] N d = ln(n) As N, the integral (and therefore the sum) become infinite. This is known as an integral comparison test.

9 8. Problem 8 See the file probem8.pdf

10 9. Problem 9 Define the Heaviside function by H(t) { 0 if t < 0 if t 0 Show, uising an ɛ δ argument that limit of H(t) as t 0 does not eist. Solution This is very similar to Problem 3. Again, one uses a proof by contradiction. Suppose the limit did eist, then there would be an L such that given an ɛ > 0, then < δ would imply H() L < ɛ. But, for any δ > 0 we can find two values such that we must have H( ) L = 0 L < ɛ and H( ) L = L < ɛ. This leads to a contradiction if ɛ > /.

11 . Problem Show that lim = 0 + Solution Since = (e ln ) = e ln, we can use the properties of the limits, and the fact that e is continuous, to show that lim 0 = lim + 0 e ln = e lim 0+ ln + To calculate the last limit, we use L Hôpital s rule. Therefore, ln lim ln = lim / = lim / 0 + / = lim = lim = e lim 0+ ln = e 0 = 0 + You can check this out on a calculator for values very close to 0.0