The following demonstrates the creation of three tables of values that might be used to determine the one-sided and two-sided limits at 0 for the indicated function.

f[x_]:=((x^2)+x)/(Sqrt[(x^3)+(x^2)])

Grid[{{TableForm@Table[{x,f[x]},{x,-.4,.4,.1}],TableForm@Table[{x,f[x]},{x,-.04,.04,.01}],
TableForm@Table[{x,f[x]},{x,-.004,.004,.001}]}},Spacings-> 2]//Quiet

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

Use commands similar to those above (cut and paste, but modifying as necessary) to find (using tables) and verify (by examing the graph) the limit value  indicated in #28, #40, and #48 on page 98 of your text.

Now let’s look at a visual demonstration of the squeeze theorem, using the application from #35 on p. 107.

f[x_]:=(x^2)Cos[20*π*x]

g[x_]:=(x^2)

h[x_]:=-(x^2)

Plot[{f[x],g[x],h[x]},{x,-1,1},PlotRange->All]

Use commands similar to those above (cut and paste, but modifying as necessary) to find (by examing the graph) the limit value  indicated in #36 on page 107 of your text.

Now let’s look at #7 on p. 117.

f[x_]:=x^3-3x+4

Plot[{f[x],6.2,5.8},{x,1,3},PlotRange->{4,8}]

To find the intersection point of the function with each of the horizontal lines, we can use the Solve command.

Solve[f[x]==6.2,x]

Solve[f[x]==5.8,x]

2.0219-2

2-1.97744

Since δ has to be the smaller of the distances from 2 to the x-coordinates of the intersection points, let’s define it accordingly.

δ=0.0219

Plot[{f[x],6.2,5.8},{x,2-δ,2+δ},PlotRange->{5.5,6.5}]

Use commands similar to those above (cut and paste, but modifying as necessary) to find (by examing the graph) the determine the solution for #8 on page 116 of your text.