1.00000000000004 10^(18)-1.00000000000001 10^(18)

10^18*(1.00000000000004-1.00000000000001 )

10^17*(10.0000000000004-10.0000000000001 )

10^16*(100.000000000004-100.000000000001 )

Plot[((x+1)-1)/x, {x,-1,1}]

Plot[((x+1)-1)/x, {x,-0.00001,0.00001}]

How do we show visually that the limit as x approaches infinity of ((2(x^2)+3x+4)/((x^2)+2x+5)) is 2?

Plot[((2(x^2)+3x+4)/((x^2)+2x+5)), {x,0,1000000}, PlotRange->All]

What do we see if the limit doesn’t exist?  Let’s try changing the (x^2) into an (x^3).

Plot[Sin[x]((2(x^3)+3x+4)/((x^2)+2x+5)), {x,0,1000000}, PlotRange->All]

f[x_] := (((((x^3+4))^2)-(x^6))/(x^3))

Plot[f[x],{x,1,1000000}]

Simplify[(((((x^3+4))^2)-(x^6))/(x^3))]

h[x_] := (((1-Cos[(x^2)]))/(x^4))

Plot[h[x],{x,-.01,.01}]

Plot[((Sin[x^2]^2)/(x^4 (1+Cos[x^2]))), {x,-.01, .01}]

j[x_] := x (Sqrt[(x^2)+4]-x)

Plot[j[x],{x,1,100000000}, PlotRange->All]

Plot[((4x)/((Sqrt[x^2+4])+x)), {x,1,100000000}(*, Axes->None*)]

a=1.1;
Plot[3x^2+1/100 Log[Abs[x-1]], {x,1-a, 1+a}]