limit of a function is giving another limit

I tried calculating the limit of the function below and matlab gave me the answer as another limit. Would i need to change the function so that i get a better answer? If I do need to change it please let me know how I could do it.

fun2 = 0.5.*(1./(y.^2)).*besselj(0,(2.*pi./lambda).*(1./y^0.5)).*log((-z_0.*y^0.5+sqrt((z_0.^2).*(y)+1))./(-z_1.*y^0.5+sqrt((z_1.^2).*y+1)));
limit(fun2,y,0)

ans =

limit((log(((y + 1)^(1/2) - y^(1/2))/((4*y + 1)^(1/2) - 2*y^(1/2)))*besselj(0, pi/(50*y^(1/2))))/y^2, y, 0)/2

Answer 1

[Sulaymon Eshkabilov]

Because Symbolic Math Toolbox could not solve your exercise and find an explcit solution formulation.

[Shubhankar Jape]

So do you know any toolbox whixh could solve this?

Answer 2

[Steven Lord]

Let's look at your function.

syms y
z_0 = 2;
z_1 = 3;
lambda = 4;
fun2 = 0.5.*(1./(y.^2)).*besselj(0,(2.*pi./lambda).*(1./y^0.5)).*log((-z_0.*y^0.5+sqrt((z_0.^2).*(y)+1))./(-z_1.*y^0.5+sqrt((z_1.^2).*y+1)));
fplot(fun2, [-1, 1])

That doesn't look all that promising in the vicinity of x = 0 for the limit to exist. Let's zoom in a bit.

figure
fplot(fun2, [-0.001 0.001])

Yeah, that really looks to me like the limit you're trying to compute does not exist which means you're encountering the first of Cleve's Golden Rules of computation.

[Shubhankar Jape]

Well yeah, but the way I thought about it was that since when y tends to go to 0, we have the natural logarithm going to 1, which leads to the whole limit going to 0.

[Steven Lord]

syms y
z_0 = 2;
z_1 = 3;
lambda = 4;
fun2 = 0.5.*(1./(y.^2)).*besselj(0,(2.*pi./lambda).*(1./y^0.5)).*log((-z_0.*y^0.5+sqrt((z_0.^2).*(y)+1))./(-z_1.*y^0.5+sqrt((z_1.^2).*y+1)))

fun2 = 

As y goes to 0, goes to Inf. How does behave as z goes to Inf? It oscillates according to the DLMF. That oscillation is multiplied by which is also going to Inf. So the appearance of the plot makes sense to me looking at your equation. If you were to evaluate fun2 for smaller and smaller values of y:

yvec = 10.^(-1:-1:-10).';
vpa(subs(fun2, y, yvec))

ans = 

This is in agreement with the plot as well.