L'hopital rule fails with limits to infinity?

$$ lim_{n to infty} frac{1 +cn^2}{(2n+3 + 2 sin n)^2} = ? $$

if I factor the $n^2$ out of denominator,

$$ lim_{n to infty} frac{ 1 + cn^2}{ n^2 ( 2 + 3n^{-1} + 2 frac{ sin n}{n} )^2}$$

And take limit directly, I get the answer as

$$ frac{c}{4}$$

However, If I apply l’hopital rule, Iget

$$ lim_{ n to infty} frac{ 2cn}{2 (2n + 3 + 2 sin n)( 2 + 2 cos n)} $$

However this new limit gives a different value than original according to wolfram.. and neither am I able to compute it by hand, what am I missing?

Some people say of limit existing and not existing, but then suppose

$$ lim_{x to 0} frac{1}{x} = infty$$

Does this limit exist? how do you define a limit to be existing as in what is sufficent condition for it