Examples

Front 1

F(x)={0 if x in Q and 1 if x in R\Q}-has no limit

Back 1

a. Take E=(1/2)
d be a positive integer
x1 in (x0-d1x0) with x1 in Q and pick x2 in (x0-d1x0) with x2 in R\Q
Then x1.x2 in D\x0 and abs(x1-x0)<d and abs(x2-x0)<d but either abs (f(x1)-L)>=E or abs(f(x2)-L)>=E
Suppose abs(f(x1)-L)=(L)<1/2, then abs(f(x2)-L=abs(1-L)>=|1-abs(L)|>1/2
Thus we have an x in D\x0 with abs(x-x0)<d but abs(F(x)-L)>=E

Front 2

G: (0,∞)-->R and g(x)=xsin(1/x)===claim lim┬(n→0)g(x)=0

Back 2

a. Let E>0 be given
Choose d=E, then d>0 and for all x in D, with |x-x0|=|x|<d we have |g(x)-0|=|xsin(1/x)|≤ |x|<d=E

Front 3

F: (0,∞)R, x0=0 and f(x)=3/x NO LIMIT

Back 3

a. Let LinR, to show L isn’t a limit of at x0, take E=1
Let d binR>0 and 0<x≤ min(d,3/(|L|+1) )
Then x in(0,d) and |f(x)-L|=|(3/x)-L|≥1
Since (3/x)>|L|+1, which implies 3/x not in (L-1,L+1)
b. Let xn=1/n. then by thrm 16 xnin D xn not =0 for all n in N but F(xn)=(3n) doesn’t converge-unbounded

Front 4

f:R->R, x->[x], f has a limit at x0 if x0 isn’t an integer

Back 4

<= suppose x0not inZ and a=[x0], then x0 in (a,a+1)
Let xn be any sequence in R\{x0} which converges to x0
Show that f(xn) converges to a, pick N such that for all n≥N we have |xn-x0|<min(x0-a,a+1-x0) for all n≥N we have xn in(a,a+1) and hence F(xn)=a
F(xn) converges to a
=> Let x0=Z then f(x0)=x0
Define xn={(x0+1/n )for n odd and (x0-1/n) for n even}
Then xn in R\{x0} and lim┬(n→∞)⁡xn =x0 since lim┬(n→∞)⁡〖1/n〗 =0
Note: f(xn)=x0 for n odd and x0-1 for n even
Hence f(x2k+1) converges to x0 while f(x2k) converges to x0-1 and hence f(xn) cant converge
Thus f doesn’t have a limit at x0

Front 5

0<c<1 is fixed
Does (nthsqrt(c) )(n in N ) converge and if so to what?

Back 5

its increasing since nthsqrt(c)-(n-1)thsqrt(c)=C^(1/n)-C^(1/n-1)=C^(1/n)(1-C^((1/n-1)-(1/n))=C^(1/n)(1-C^(1/n(n-1))>0 since C^(1/n(n-1))<1 since C<1.
nthsqrtC is bounded since 0<nthsqrtC<1 fo all n. hence by thrm nthsqrtC converges.
convege t0?

Let A be lin(nthsqrt(c))(n in N).
-From theorem 14, lim(2nthsqrt(c)) = A
-On the other hand, (2nthsqrt(c))= sqrt(nthsqrt(c)), which shows lim(2nthsqrt(c))= sqrtA
-Hence, sqrtA=A, meaning A= 0 or 1
A cannot be 0 since A is the supremeum if nthsqrt(c) is contained in (0,1)