M. Hirnyk!
Of course, the eliminate command is not always able to find an explicit equation relating B and A. In this case, the problem can be solved numerically by the following scheme (the simplest way):

1) Construct the plot of the function  B(t)  and on it we find the range in which the function is monotonic.
2) For each value of  B  through some step find the corresponding value of t .

3) Substitute these  t in A (t) .
4) Using these lists we build the plot A (B) .

Solution for your example:

plot(t^5-sin(t)-2, t=0..2, labels=[t,B]);

T:=[seq(fsolve(B = t^5-sin(t)-2, t=0.63..2), B=-2.4..29, 0.1)];

X:=[seq(B, B=-2.4..29, 0.1)];
Y:=[seq(eval( t^3-exp(t)), t in T)];

plot(X, Y, labels=[B,A]);

You have 2 functions  A=A(t)  and B=B(t) .To plot A in terms of B you should to eliminate the variable  t . 

An example:

F := eliminate({A = t^3-t, B = t^3-t^2}, t);

plots[implicitplot](op(F[2]), B = -1 .. 1, A = -1 .. 1, scaling = constrained, numpoints = 50000);

If I understand your question, you need to build a point plot of the function  x->x/(x+1)  for a sets of points  0.5, 0.9, 0.99, 0.999  and  1.001, 1.01, 1.1, 1.5  to illustrate the concept of limit of a function when the variable tends to 1. This can be done as follows

X1, X2 := [.5, .9, .99, .999], [1.001, 1.01, 1.1, 1.5]:

f := x->x/(x+1):

Y1, Y2 := map(f, X1), map(f, X2):

plots[display](plot(X1, Y1, style = point, symbol = solidcircle, color = blue, symbolsize = 8), plot(X2, Y2, style = point, symbol = solidcircle, color = red, symbolsize = 8), view = [0 .. 2, 0 .. 1], scaling = constrained);

restart;

f := -k*x1(t)+k*(x2(t)-x1(t))-lambda*(diff(x1(t), t)) ;

g := -k*(x2(t)-x1(t));

eq1 := m*(diff(x1(t), t, t))-f = 0;

eq2 := m*(diff(x2(t), t, t))-g = 0;

fcns := {x1(t), x2(t)};

ICS := {x1(0) = 1, x2(0) = 0, D(x1)(0) = 0, D(x2)(0) = 0};

sys := {eq1, eq2}; m := 1; k := 1; lambda := 1;

sysdiff := sys union ICS;

Sol := dsolve(sysdiff, fcns, numeric);

plots[odeplot](Sol, [[t,x1(t)],[t,x2(t)]], t=0..30, color=[red, blue], numpoints=5000, thickness=2);

restart :

f := -k*x1(t)+k2*(x2(t)-x1(t))-lambda*(diff(x1(t), t)):

g := -k2*(x2(t)-x1(t)): fcns := x1(t), x2(t):

eq1 := m*(diff(x1(t), t, t))-f = 0:

eq2 := m*(diff(x1(t), t, t))-g = 0:

ICS := {x1(0) = 1, x2(0) = 0, D(x1)(0) = 0}:

sys := {eq1, eq2}:

m := 1: k2 := 1: k1 := 1: lambda := 1: k:=1:

sysdiff := `union`(sys, ICS):

Sol_N := dsolve(sysdiff, {fcns}, numeric):

plots[odeplot](Sol_N , [x1(t),x2(t)],0..100, numpoints=5000, thickness=2, view=[-0.5..0.5, -0.5..0.5]);

a[n] is `in` S;

plot([[4*cos(t), 3*sin(t), t=0..2*Pi], [2*cos(t), 3*sin(t), t=0..2*Pi]], scaling=constrained, color=[red, blue]);

f:=x->x^2+2*x-3:
D(f)(2);

                  6

Simplification.mw

restart;

L:=[]:

 for A  from .1 by .01 to 3 do

  for B  from .1 by .01 to 3 do

  l1:=(exp(B)-A)-1:

  l2:=(exp(B)-A/2)-1.5:

  if l1>0 and l2>0 then L:=[op(L), [A,B]]: break: end if:

  end do:

 end do:

L; 

In total there are 192 solutions:

with(combinat):

S:=permute(9): L:=[]:

for a in S do

if  a[9] = add(a[2*k-1]/a[2*k], k= 1..4) then L:=[op(L), a]: fi:

od:

nops(L);

                                               192

The first solution in the list  L  is Carl's solution, the last is

It is interesting that  there is no solution in which all the fractions are irreducible.

There are several solutions.

solve({c2+vc = .34, 

c2+c3 = .519,

t8+c3-c4-vc = .132,

t1+t5-c2+0.0435=0., 

t1-t2-c2+c3-1.334*vc = 0.,

t2+t6-c3-c4+0.0435+1.334*vc=0.,

-112.5*t1+112.5*t5+300*c2^2+1000*vc = 38.25, 

112.5*t1-112.5*t2-300*c2^2+300*c3-300*c3^2-2000*vc+23.4 = 0.,

-112.8*t8+112.5*t2-112.5*t6+300*c4^2-300*c3+300*c3^2+1000*vc+14.85 = 0.}, 

{t1,t2,t5,t6,t8,c2,c3,c4,vc});

{c2 = .3099822380, c3 = .2090177620, c4 = .2431877898e-1, t1 = .2247720450, t2 = .8376387448e-1, t5 = .4171019302e-1, t6 = .6602897199e-1, t8 = -.2268122102e-1, vc = .3001776198e-1}, {c2 = 3.167226095, c3 = -2.648226095, c4 = .2431877898e-1, t1 = 2.201508587, t2 = .1575760075, t5 = .9222175084, t6 = .9465362873, t8 = -.2268122102e-1, vc = -2.827226095}, {c2 = .3099822380, c3 = .2090177620, c4 = .7266812210, t1 = .2247720450, t2 = .8376387448e-1, t5 = .4171019302e-1, t6 = .7683914140, t8 = .6796812210, vc = .3001776198e-1}, {c2 = 3.167226095, c3 = -2.648226095, c4 = .7266812210, t1 = 2.201508587, t2 = .1575760075, t5 = .9222175084, t6 = 1.648898729, t8 = .6796812210, vc = -2.827226095}

The original code has been executed in Maple 12 (classical worksheet). Found mistake in Dist2 (bracket in the wrong place).
  
Here's another variant of the code:

M1 := [-2, 3, -4]: M2 := [4, -1, 3]:

A := [-t+3, 2*t-4, t+2]:

B := [s+4, 2*s, 2*s-1]:

Dist1 := sqrt((M1[1]-A[1])^2+(M1[2]-A[2])^2+(M1[3]-A[3])^2)+sqrt((B[1]-A[1])^2+(B[2]-A[2])^2+(B[3]-A[3])^2)+sqrt((B[1]-M2[1])^2+(B[2]-M2[2])^2+(B[3]-M2[3])^2):

Dist2 := sqrt((M1[1]-B[1])^2+(M1[2]-B[2])^2+(M1[3]-B[3])^2)+sqrt((B[1]-A[1])^2+(B[2]-A[2])^2+(B[3]-A[3])^2)+sqrt((A[1]-M2[1])^2+(A[2]-M2[2])^2+(A[3]-M2[3])^2):

Optimization[Minimize](Dist1); Optimization[Minimize](Dist2);

min(%[1], %%[1]);

[17.8369577415605676,

[s = 0.602330835119798569, t = 1.52243814242405584]]

[14.4132655077310190,

[s = -0.189757117490280680, t = 0.845165089700380400]]

14.4132655077310190

This is quite simply:

M1:=[-2,3,-4]:  M2:=[4,-1,3]:

A:=[-t+3,2*t-4,t+2]:  B:=[s+4,2*s,2*s-1]:

Dist1:=sqrt((M1[1]-A[1])^2+(M1[2]-A[2])^2+(M1[3]-A[3])^2)+sqrt((B[1]-A[1])^2+(B[2]-A[2])^2+(B[3]-A[3])^2)+sqrt((B[1]-M2[1])^2+(B[2]-M2[2])^2+(B[3]-M2[3])^2):

Dist2:=sqrt((M1[1]-B[1])^2+(M1[2]-B[2])^2+(M1[3]-B[3])^2)+sqrt((B[1]-A[1])^2+(B[2]-A[2])^2)+(B[3]-A[3])^2+sqrt((A[1]-M2[1])^2+(A[2]-M2[2])^2+(A[3]-M2[3])^2):

evalf(minimize(Dist1));  evalf(minimize(Dist2));

min(%, %%); 

                                                                         21.59170144

                                                                         16.24881940

                                                                         16.24881940

1) If you consider the function y=psi(x)  in the real range, then you should write surd(1-x, 3)  instead  (1-x)^(1/3). Else for  (1-x)^(1/3)  if  x>1 , you get complex values.

2) By the code  q:=y->fsolve(y=psi(x), x)  you are looking the function inverse to the function y=psi(x) . For correct solution of the problem the function y=psi(x) should be monotonic in the corresponding range. Such ranges simpliest can be found using the plot of the function. Function  y=psi(x)  in the entire domain does not have an inverse function.

V:=1.8e-5:  R:=8.314:  T:=298:  k:=0.841:  G:=1e6:

psi:=unapply((ln(x)-x+k*(1-x)^2+(V*G)/(R*T)*(surd(1-x,3)-(1-x)/2))*(R*T/V), x);

plot(psi(x), x=0..3, title="The plot of function y=psi(x)");

q:=y->fsolve(psi(x)=y, x):

plot('q'(y), y=-.9e9..-.1.4e9);