How to find a limit without syms and limit function

Question

Florin Florin 2019-5-25

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https://ww2.mathworks.cn/matlabcentral/answers/463969-how-to-find-a-limit-without-syms-and-limit-function

编辑： John D'Errico 2021-10-6

采纳的回答： Star Strider

Let's take the limit

. How can i calculate it without using syms and matlab's function limit?

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Answer 1

Star Strider 2019-5-25

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编辑：Star Strider 2019-5-25

在 MATLAB Online 中打开

Crude but effective (for this function, may not be universally applicable):

fcn = @(x) (x.^3 - 1) ./ (x - 1);
x = 1;
lm = fcn(x-1E-15)
lm =
     3

Experiment to get the result you want.

EDIT —

Another option is to use a simple numerical derivative:

dfdx = @(f,x) (f(x + 1E-8) - f(x)) ./ 1E-8;
fcnn = @(x) x.^3 - 1;
fcnd = @(x) x - 1;
xv = 1;
Lm = dfdx(fcnn,xv) ./ dfdx(fcnd,xv)

producing:

Lm =
3

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Gustav Garpebo 2021-10-6

Can you pleasae explain the terms? xv, fcnn, fcnd etc

Star Strider 2021-10-6

@Gustav Garpebo — Sure! (I probably should have explained those originally, describing them in comments, although they were clear in the context 2½ years ago.)

The forward-difference derivative ‘dfdx’ function requires a function handle (first argument) and a value of ‘x’ at which the function is evaluated (second argument), and since the function is being evaluated at 1 that is what ‘xv’ is assigned to be. The two other functions, ‘fcnn’ and ‘fcnd’ are the numerator and denominator of the original function, respectively. The rest is straightforward.

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Answer 2

John D'Errico 2021-10-6

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https://ww2.mathworks.cn/matlabcentral/answers/463969-how-to-find-a-limit-without-syms-and-limit-function#answer_802651

编辑：John D'Errico 2021-10-6

在 MATLAB Online 中打开

You can use my limest function. It is on the file exchange.

>> fun= @(x) (x.^3 - 1)./(x-1)
fun =
  function_handle with value:
    @(x)(x.^3-1)./(x-1)

Now use limest. It even provides an estimate of how well it thinks that limit is known.

[L,errest] = limest(fun,1)
L =
                         3
errest =
      2.20957326622612e-14

Is that correct? l'hopital would tell me of course. Thus, if I differentiate the numerator and the demoninator, we would have 3^x^2/1. At x==1, that is 3.

The symbolic toolbox would agree, but you don't want to see that.

syms X
F = (x^3-1)/(x-1)
limit(F,1)
ans =
3

But we can still use the symbolic TB, without use of limit, just using l'hopital...

subs(diff(X^3-1),X,1)/subs(diff(X-1),X,1)
ans =
3

As expected, it returns 3 as the desired limit.

You can find limest on the file exchange, here:

https://www.mathworks.com/matlabcentral/fileexchange/20058-adaptive-numerical-limit-and-residue-estimation?s_tid=srchtitle

LIMEST uses an adaptive, multi-order Richardson extrapolation scheme, modified to provide also an estimate of the uncertainty at the extrapolation point, all of my invention.)