The problem will disappear if you write:

f := x->piecewise(x <= 1, a*x^2+b, 1 <= x, 1/x):

limit(diff(f(x), x), x = 1, right);

limit(diff(f(x), x), x = 1, left);

                                 -1

                                 2 a

Cause of error - the unknown value of the function  f(1)  to calculate the right derivative  at  x=1 .

I do not understand the meaning of the problem. A curve is a set of points, depending on one parameter. In your expressions, there are several parameters. 

with(LinearAlgebra):

add(add(A[i,j]*B[i,j], i=1..RowDimension(A)), j=1..ColumnDimension(A));

S1 := seq(plot3d(0, x = -11 .. h, y = 0 .. sqrt(121-x^2), style = surface), h = -11 .. 11, .2):

S2 := seq(plot3d([x, y, y*sqrt(3)], x = -11 .. h, y = 0 .. (1/2)*sqrt(121-x^2), style = surface), h = -11 .. 11, .2):

S3 := seq(plot3d([x, y, (sqrt(121-x^2)-y)*sqrt(3)], x = -11 .. h, y = (1/2)*sqrt(121-x^2) .. sqrt(121-x^2), style = surface), h = -11 .. 11, .2):

S4 := seq(plottools[polygon]([[x, 0, 0], [x, (1/2)*sqrt(121-x^2), (1/2)*sqrt(121-x^2)*sqrt(3)], [x, sqrt(121-x^2), 0]], color = pink), x = -11 .. 11, .2): C := plots[spacecurve]([11*cos(t), 11*sin(t), 0], t = 0 .. Pi, color = black):

S := seq(plots[display](C, S1[k], S2[k], S3[k], S4[k]), k = 1 .. 111):

plots[display](S, insequence = true, scaling = constrained, axes = normal, lightmodel = light4, view = [-12.5 .. 12.5, -2.5 .. 12.5, -1.5 .. 11.5], orientation = [-12, 67]);

restart;

L:=[]: Set:={$0..9}:

for H in {1,2,3} do

for U in Set minus {H} do

for N in Set minus {H,U} do

for S in Set minus {0,H,U,N} do

for E in Set minus {H,U,N,S} do

for T in Set minus {H,U,N,S,E} do

for R in Set minus {H,U,N,S,E,T} do

for o in Set minus {H,U,N,S,E,T,R} do

for G in Set minus {H,U,N,S,E,T,R,o} do

a:=H*10^5+U*10^4+N*10^3+S*10^2+E*10+N: b:=S*10^5+T*10^4+R*10^3+o*10^2+N*10+G:

if a*3=b then L:=[op(L), [HUNSEN=a, STRONG=b]] fi:

od: od: od: od: od: od: od: od: od:

            [[HUNSEN = 126356, STRONG = 379068], [HUNSEN = 318928, STRONG = 956784]]

The method of solution is similar to one in  http://www.mapleprimes.com/questions/151971-Number-Theory

Obviously, the vertices of the alternating triangle are

A := [x, 0, 0]: B := [x, sqrt(121-x^2), 0]: C := [x, (1/2)*sqrt(121-x^2), (1/2)*sqrt(121-x^2)*sqrt(3)]: 

The code for plotting:

S1 := plot3d(0, x = -11 .. 11, y = 0 .. sqrt(121-x^2), style = surface):  # Lower face of the solid 

S2 := plot3d([x, y, y*sqrt(3)], x = -11 .. 11, y = 0 .. (1/2)*sqrt(121-x^2), style = surface):  # Left face of the solid

S3 := plot3d([x, y, (sqrt(121-x^2)-y)*sqrt(3)], x = -11 .. 11, y = (1/2)*sqrt(121-x^2) .. sqrt(121-x^2), style = surface):  #Right face of the solid

plots[display](S1, S2, S3, scaling = constrained, axes = normal, lightmodel = light4, view = [-12.5 .. 12.5, -2.5 .. 12.5, -1.5 .. 11.5], orientation = [-12, 67]);

Now your body is closed from all sides:

A := plot3d([2*t^3-t^4, (t+2)*cos(theta)-2, (t+2)*sin(theta)], t = 0 .. 2, theta = 0 .. 2*Pi):

B := plots:-spacecurve([t, -2, 0], t = -1 .. 3, color = red, thickness = 2):

C := plot3d([0, -2+r*cos(phi), r*sin(phi)], r = 2 .. 4, phi = 0 .. 2*Pi):

plots:-display(A, B, C, axes = normal, scaling=constrained, view = [-1.4 .. 5.4, -7.4 .. 3.4, -4.4 .. 4.4]);

Code of  Preben Alsholm  was used.

Without plots and plottools:

piecewise(seq(op([x>-7+2*n and x<-7+2*(n+1), exp(-x-6+2*n)]), n=0..7)):

plot(%, x=-7..9, -1..3.5, thickness=2, scaling=constrained, discont=true);

If a function was specified by an expression, as the questioner did, then

f(x):=((x^2-x-3)/(x^2+1))*(x^2+x+1):

diff(f(x), x);

simplify(%);

The result is the same.

restart;

S:={$0..9}: L:=[]:

for C in (S minus {0}) do

for N in (S minus {C}) do

for R in (S minus {C,N}) do

for P in (S minus {C,N,R}) do

for K in (S minus {0,C,N,R,P}) do

for H in (S minus {C,N,R,P,K}) do

for M in (S minus {C,N,R,P,K,H}) do

for E in (S minus {C,N,R,P,K,H,M}) do

a:=C*10^3+N*10^2+R*10+P:  b:=C*10^2+P*10+P:  c:=K*10^4+H*10^3+M*10^2+E*10+R:

if a + b = c  then  L:=[op(L), [CNRP=a, CPP=b, KHMER=c]]  fi:

od: od: od: od: od: od: od: od:

op(L);

Received a unique solution.               

The first term is already linear in  f .

expression := diff(f(r), r$2) + exp(-f(r))*(1 - g(r))^2:

op(1,expression)+subs({f=f(r), g=g(r)},mtaylor(eval(op(2,expression), {f(r)=f, g(r)=g}), [f, g], 2));

Let  in the pocket  a is the number of  50c coins,  b  is the number of  20c coins,  c  is the number of  10c coins,  d  is the number of  5c coins. It is evident that  a<=1,  b<=4,  c<=9,  d<=19

M:=[]:

for a from 0 to 1 do

for b from 0 to 4 do

for c from 0 to 9 do

for d from 0 to 19 do

if 50*a+20*b+10*c+5*d>100 and convert([seq(seq(seq(seq(50*i+20*j+10*k+5*l<>100, i=0..a),j=0..b), k=0..c), l=0..d)], `and`) then M:=[op(M),[a,b,c,d]] fi:

od: od: od: od:

M;

max(seq(50*M[i,1]+20*M[i,2]+10*M[i,3]+5*M[i,4], i=1..nops(M)));

Another way with two arguments:

seq(a[4 - i], i=1 .. 3);

Carl's  method does not work for me (Maple 16).

The function  x->x^x  is defined only for  x>0 .

Your example A can be easily solved by hand, if we note that this limit is the derivative of the function  f  at  the point  x=n :

f:=x->x^x:

Limit(((n+k)^(n+k)-n^n)/k, k=0)=D(f)(n);

Direct calculation of the limit with Maple give the same result:

limit(((n+k)^(n+k)-n^n)/k, k=0);

If you need in manual solution or want to verify Carl's result you can rewrite the sum to reduce it to simple definite integral:

sum(1/(n*(1+k/n)), k = 1 .. n):

limit(%, n = infinity) = Int(1/(n+1), n = 0 .. 1);

int(1/(n+1), n = 0 .. 1);