Section 1.6: Limit Laws
Andy Long
Spring, 2016

(* Section 1.6: Calculating Limits Using Limit Laws *)
Clear[a,b,f,g,PlotA,PlotB];

(* Example 3, p. 65: *)
f[x_]:=(x^2-1)/(x-1)
a=1
b=Limit[f[x],x->a]
R={{-.1,2},{-.1,3}};
PlotA=Plot[f[x],{x,a-1,a+1},PlotRange->R];
PlotB=ListPlot[{{a,0},{0,b},{a,b}},PlotMarkers->{"☹"}];
c=.9
PlotC=ListPlot[{{c,0},{0,f[c]},{c,f[c]}},PlotMarkers->{"☺"}];
Show[PlotA,PlotB,PlotC,PlotRange->R]

(* Now let's see that in action: *)
f[x_]:=(x^2-1)/(x-1)
Manipulate[
Module[
{},
PlotC=ListPlot[{{x,0},{0,f[x]},{x,f[x]}},PlotMarkers->{"☺"}];Show[PlotA,PlotB,PlotC,PlotRange->R]
]
,{x,0,2}
]

(* pp.69-71,#2,7,10,25,33,47,54 *)

(* #7 *)
f[x_]:=(1+x^(1/3))(2-6x^2+x^3)
Plot[f[x],{x,7,9}]
Limit[f[x],x-> 8]
Limit[(1+x^(1/3)),x-> 8]
Limit[1,x-> 8]+Limit[x^(1/3),x-> 8]
Limit[(2-6x^2+x^3),x-> 8]
Limit[2,x-> 8]-Limit[6x^2,x-> 8]+Limit[x^3,x-> 8]

(* #10 *)
f[x_]:=(x^2+x-6)/(x-2)
g[x_]:=x+3
Limit[f[x],x-> 2]
Limit[g[x],x-> 2]

(* #25: multiply by the conjugate *)
f[t_]:=(Sqrt[1+t]-Sqrt[1-t])/t
Limit[f[t],t-> 0]
(* by multiplying by the conjugate, we eliminate the problem of 0/0: *)
g[t_]:=((1+t)-(1-t))/(t*(Sqrt[1+t]+Sqrt[1-t]))
g[t]
Limit[g[t],t-> 0]

(* #33 *)
f[x_]:=x/(Sqrt[1+3x]-1)
Plot[f[x],{x,-.3,.3}]
(* From the left: *)
list={-.3,-.1,-.01,-.001,-.0001}
 TableForm[Table[{x,f[x]},{x,list}]]
(* From the right: *)
 TableForm[Table[{x,f[x]},{x,-list}]]
Limit[f[x],x-> 0]
g[x_]:=(x*(Sqrt[1+3x]+1))/((1+3x)-1)
g[x]
Limit[g[x],x-> 0]

(* #47 *)
f[x_]:=Sign[x]
Plot[f[x],{x,-2,2}]
f[0]
Limit[f[x],x -> 0] (* a lie! *)
Limit[f[x],x -> 0,Direction->1]
Limit[f[x],x -> 0,Direction->-1]

(* #54 *)
L[v_]:=Lo*Sqrt[1-v^2/c^2]
L[v]
Limit[L[v],v -> c,Direction->1]

(* #63: *)
Manipulate[
Plot[(3x^2+a*x+a+3)/(x^2+x-2),{x,-3,-1},PlotRange->{-4,4}],{a,10,20}
]