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> limit(sin(x)/x,x=0);
Notice that the first arguement in the parentheses is the expression that you are taking the limit of; the second is the c-value that x is approaching. As in the previous example, you can type the expression directly into the limit command. Another approach is to first define the function in Maple and use the function name within the limit command.
> limit(x^2+8*sin(-3*x*Pi), x=1/2);
> f:=x->(cos(x)-1)/x;
> limit(f(x),x=0);
If the limit exists, Maple can usually
find it. In cases where the limit doesn't exist, such as
or
where , Maple gives the
answer undefined
. Look at the plots of the functions to see
why the limits are undefined.
> limit(1/x,x=0);
> plot(1/x,x=-5..5,y=-10..10);
> p := piecewise(x<0,-x^2+4,x>=0,x-2);
> limit(p,x=0);
> plot(p,x=-5..5);
Another reason the limit does not exist at x0 is because
of oscillations. The limit as x approaches 0 of is undefined. To get a better idea of why, look at the plot. When
evaluating the limit using Maple, the result is the range
-1..1
. When the limit doesn't exist, but the expression or
function is bounded, this is the type of answer that Maple will give.
> plot(cos(1/x),x=-2..2);
> limit(cos(1/x),x=0);
Maple also can find one-sided limits. Suppose that you want to find the limit as x approaches 0 from the right for the following function:
> f:=x->x*sin(1/x);
> limit(f(x),x=0,right);
Similarly, you can find the limit of f(x) as x approaches 0 from the left.
The derivative of a function f(x), often written f'(x), is defined by the following limit.
When using the definition to compute a derivative in Maple, it is easiest to first define the function f. You can then find the difference quotient of f and take the limit as h approaches 0 in either one or two steps. Both examples are shown below.
> f := x -> x^2+3*x+5;
> DQ:=(f(x+h)-f(x))/h;
> derivative:=limit(DQ,h=0);
> derivative:=limit((f(x+h)-f(x))/h,h=0);
To evaluate the derivative at a given x value, you can use one of two methods. Both are given below.
> subs(x=1,derivative);
> limit ((f(1+h)-f(1))/h,h=0);
Christine M Palmer