Limits problem 08 (Squeeze).mws

1. The Squeeze Theorem for Limits.

One way in which limits are often established is by the Squeeze Theorem.  If f(x) <= g(x) and g(x) <= h(x) for all x in an open interval that contains a (except possibly at a ) and limit(f(x),x = a) = limit(h(x),x = a) = L , then limit(g(x),x = a) = L .  We can use this theorem to show that limit(x^2*cos(20*Pi*x),x = 0) = 0 , and to illustrate this fact by graphing. Notice that the "amplitude" of this function is x^2 . The curve y = x^2 lies above the graph of y = x^2*cos(20*Pi*x) , while the curve y = -x^2 lies below the graph.

> f:=x->-x^2: g:=x->x^2*cos(20*Pi*x): h:=x->x^2:

> plot([g(x),f(x),h(x)],x=-Pi..Pi,color=[red,green,blue],thickness=[1,3,3]);

[Maple Plot]

> plot([g(x),f(x),h(x)],x=-Pi/4..Pi/4,color=[red,green,blue],thickness=[1,3,3]);

[Maple Plot]

> plot([g(x),f(x),h(x)],x=-Pi/20..Pi/20,color=[red,green,blue],thickness=[1,3,3]);

[Maple Plot]

> plot([g(x),f(x),h(x)],x=-Pi/200..Pi/200,color=[red,green,blue],thickness=[1,3,3]);

[Maple Plot]

To establish the limit using the squeeze theorem, we note that  -1 <= cos(20*Pi*x) and cos(20*Pi*x) <= 1 , so -x^2 <= x^2*cos(20*Pi*x) and x^2*cos(20*Pi*x) <= x^2 . We know that  limit(x^2,x = 0) = 0 andlimit(-x^2,x = 0) = 0 , so it follows from the squeeze theorem that limit(x^2*cos(20*Pi*x),x = 0) = 0

Submission:

(a) Use the squeeze theorem to show that limit(sqrt(x^3+x^2)*sin(Pi/x),x = 0) = 0 .

(b) Graph your functions f(x) , g(x) , and h(x) on the same set of axes.

Submission worksheet: