How can i use L'hospital rule i.e sin(x)/x for x tends to zero?

I have a simple program

n=[0:7] x=sin((pi*n)/2)./(pi*n)

which returns,

n =

     0     1     2     3     4     5     6     7

x =

       NaN    0.3183    0.0000   -0.1061   -0.0000    0.0637    0.0000   -0.0455

Only the problem is that,according to L'Hospital rule,value of x for n=0 should be 0.5 instead getting NaN.. How can i get this? Is there any command?

Answer 1

[Star Strider]

By convention, 0/0 evaluates to NaN.

To be absolutely formal (if a bit kludgy) about using L’Hospital's rule in this situation:

fderiv = @(f,x) (f(x+eps)-f(x))/eps;                    % Derivative of a function
xderiv = @(x) ((x+eps)-x)/eps;                          % Derivative of an expression
for n = 0:7
    x(n+1)=sin((pi*n)/2)./(pi*n);
    if isnan( x(n+1) )                                  % Test for ‘NaN’
        x(n+1) = fderiv(@sin, pi*n/2)./(xderiv(pi*n));  % L'Hospital's Rule
    end
end

There are better and more numerically stable derivative expressions (such as the five-point algorithm), but this will do for an illustration.

Also, using L'Hospital's rule,

x=sin((pi*n)/2)./(pi*n)

will be 1 at n=0, however:

x=(sin(pi*n)/2)./(pi*n)

with the fderiv call then becoming:

x(n+1) = (fderiv(@sin, pi*n)/2)./(xderiv(pi*n));

will be 0.5 for n=0.

[Sudhir Dhadge]

Thanks for the answer..And also for the correction...I am very happy..

[Star Strider]

It is my pleasure to help!

Answer 2

[Wayne King]

You can do this:

    x = -5:0.001:5;
    y = sin(x)./x;
    plot(x,y)

Since the increment in the x-variable is 0.001 and the range is 10, the total number of x-samples is 10^4, 0 will be at index 5001. Look at the y-values in a neighborhood of 0.

y(4950:5050)

You can see that the left-hand and right-hand limits are 1.

If you have the Symbolic Toolbox.

ezplot('sin(x)/x',[-5,5])