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
$\endgroup$ 131 Answer
$\begingroup$The rule of L'Hospital states that the limit of $\dfrac fg$ equals that of $\dfrac{f'}{g'}$ if the latter exists. You precisely found a case where this does not hold.
We can simplify the example as
$$\lim_{n\to\infty}\frac{n+\sin n}n=1$$
but
$$\lim_{n\to\infty}\frac{1+\cos n}1$$ is undefined.
$\endgroup$ 10
