We can easily prove by hand even an more general result, that for any number of roots in the first term (starting from 2) the limit is 1/2.
We denote by  a[n]  the expression with  n  roots. For example  a[3]=sqrt(x+sqrt(x+sqrt(x)))

It's obvious that  for any  n>=2  we have   
a[n]=sqrt(x+a[n-1]) ,  limit(a[n]/sqrt(x), x=infinity)=1,  limit(a[n]/x, x=infinity)=0

Below we prove that for any  n>=2  the result holds  limit(a[n]-sqrt(x), x=infinity) = 1/2 

We denote  A=sqrt(x+a[n-1]) - sqrt(x) ,  B=sqrt(x+a[n-1]) + sqrt(x)

For the proof, it is sufficient to make only 3 steps: multiply and divide  A  by  B , simplify in the numerator using the difference formula for the squares, and then divide the numerator and the denominator by  sqrt(x)  (below is an illustration of these steps in Maple) :

restart;
A:=sqrt(x+a[n-1])-sqrt(x);
B:=sqrt(x+a[n-1])+sqrt(x);
C:=expand(A*B)/B;
``(expand(combine(numer(C)/sqrt(x))))/map(p->expand(combine(p/sqrt(x))),denom(C)) assuming x>0;
                                

Using the above limits, we get the result  1/2 

For those who want to check the result in Maple (without any steps) , here's a procedure that allows you to automatically generate the expression  a[n]  for any  n :

restart;
a:=proc(n)
if n=1 then return sqrt(x) else
sqrt(x+a(n-1)) fi;
end proc:

Examples of use:

seq(a(n), n=[1,4,10]);

The answer is very simple - your inequality in addition to the basic unknown  r  contains several parameters  (lambda, M) and Maple simply does not know how to solve such inequalities.

Here is a much simpler example that Maple also refuses to solve:

solve(sqrt(a*x+1) < 1, x);
     Warning, solutions may have been lost

This simple example, we can still solve, if to use  assuming  option:

solve(sqrt(a*x+1) < 1, x) assuming a>0;
solve(sqrt(a*x+1) < 1, x) assuming a<0;

But for your inequality and it does not help.

I think first you need to specify the values of the parameters  lambda  and  M .

See  ?plot,axis  help page for this.

Do you just want to check that a number  a  is the root of your equation? In this case, do 

simplify(eval(F(X)=0, X=a));

Sometimes other commands are required instead of  simplify . Then specify the exact form of your equation and the value of a .
 

restart;
with(Physics):
Setup(mathematicalnotation = true);                
Setup(signature = `+---`);
ds2 := ((x^2-y^2)*cos(2*u)+2*x*y*sin(2*u))*(du^2)-2*x*(dv^2)-dx^2-dy^2;
Setup(coordinatesystems=(Z=[u, v, x, y]), metric = ds2);
 

Edit.

[{1, 2}, {2, 3}] minus~ [{2}$2];
                                                      [{1}, {3}]

A more traditional and programmatic way of obtaining the values of a function obtained as a result of solving an equation, etc., is to use  eval  command. See the same example as tomleslie's one:

Download eval.mw

See  question_(2)_new.mw

ode:= diff(y(x),x)=2*x:
plots:-display(DEtools:-DEplot(ode,y(x),x = -2 .. 2,y = -2 .. 2, [[0.1,0]],
               labels=["",""],
               linecolour = red,
               color = blue,
               'arrows' = 'medium',
               axesfont=['Times', 'bold', 12]
               ), axis=[tickmarks=[color=red]]);

For your first example, everything is the same:

restart;
plots:-display(plot(sin(x),x=-Pi..Pi, color=blue),axis=[tickmarks=[color=red]]);
 

Here is an example of solving the problem. The method is of a general nature and can be used for any curves given by parametric equations. In the example, the red curve is a broken line, and the blue one is a straight line.

restart;
d:=(P,Q)->sqrt((P[1]-Q[1])^2+(P[2]-Q[2])^2);
P:=piecewise(s>=0 and s<0.3,[s,s/3+0.3],s>=0.3 and s<0.5,[s,-s+0.7],s>=0.5 and s<0.7,[s,7/2*s-1.55],s>=0.7 and s<=1,[s,-s/3+17/15]);
Q:=[t,t];
A:=plot(piecewise(seq(op([op(P)[2*n-1],op(P)[2*n][2]]), n=1..nops([op(P)])/2)), s=0..1,color=red, thickness=3):
B:=plot(t, t=0..1, color=blue, thickness=3):
epsilon:=0.3;
C:=plots:-inequal({d(P,Q)<=epsilon}, s=0..1, t=0..1, optionsfeasible = [color = "WhiteSmoke"], optionsexcluded = [color = "DarkGrey"], view=[0..1,0..1]):
plots:-display(<plots:-display(A,B) | C>, scaling=constrained); # This is "free space diagram"

If the curve is given by a conventional equation, for example  y=f(x) , so you must specify it as the list  [x, f(x)]

FD.mw

Edit.

Try  
A.C;  # dot product 
# or  
expand(A.C); 

Type this ( . ) as an ordinary dot on the keyboard. I use 1d-math input. In 2d-math input (which you use) , this will look like a fat dot in the middle.

For real domain use  discont  command instead of  singular  (the output is empty set):

discont(ln(y^2+1), y);
                                           { }

You are not right. Maple calculates everything correctly. Should be

combinat:-numbcomb(n, m) = binomial(n, m)  

See help on this command.

For example  combinat:-numbcomb(7, 3) = (7*6*5)/3! = 35

The mapping of the square  [0,1] x [0,1]  with 2d-grid:

restart;
f:=(r,t)->[r*exp(t),r*exp(-t)];
S:=plot([seq([x,t,t=0..1],x=0..1,0.1),seq([t,y,t=0..1],y=0..1,0.1)], color=red, size=[200,200]):
F:=plottools:-transform(f):
T:=plots:-display(F(S), size=[400,400]):
plots:-display(<S | T>, scaling=constrained);

                       

Edit.
                        

Try use inert multiplication for this:

A:=x%*y=y%*x + 2*z;
expand(subs(A, w*x%*y));