Question

lim ∂->∞ ∂^2/(1-cos∂)

Answer 1

you killed the previous question too quickly.

you can't reopen it but you established that

\(\lim\limits_{x \to \infty} \cos \left( \frac{1}{x} \right)\)

\(= \cos \left( \lim\limits_{x \to \infty} \frac{1}{x} \right) = \cos 0 = 1\), because cosine is a continuous function

you can use that here

Answer 2

\(\color{black}{\displaystyle \lim _{x\to \infty }\frac{x^2}{1-\cos x}=\infty. }\)

You know that \(\color{black}{\displaystyle \cos x }\) is at least \(\color{black}{\displaystyle -1 }\) and at most \(\color{black}{\displaystyle 1 }\).

When \(\color{black}{\displaystyle \cos x=1 }\), you have (so to speak)
\(\color{black}{\displaystyle \infty^2/0=\infty^3=\infty }\)
(Well, not like that, but you know it then diverges to \(\color{black}{\displaystyle \infty}\).)

When \(\color{black}{\displaystyle \cos x }\) is any value \(\color{black}{\displaystyle \varepsilon }\), such that \(\color{black}{\displaystyle \{\varepsilon:\varepsilon\in[-1,1]~~{\rm and}~~\varepsilon\ne0\}}\)
(I've already excluded \(\color{black}{\displaystyle \varepsilon=0 }\) by excluding \(\color{black}{\displaystyle \cos x=1 }\))
\(\color{black}{\displaystyle \bf THEN{\tiny~}}\) you have the following:

\(\color{black}{\displaystyle \lim _{x\to \infty }\frac{x^2}{\varepsilon}=\frac{1}{\varepsilon} \lim _{x\to \infty }x^2=\infty}\).

Answer 3

This is all just common sense, really ... all I did is just speak it out.

Answer 4

A more interesting case though would be when \(\color{black}{\displaystyle x\to0}\).

\(\color{black}{\displaystyle \lim _{x\to 0 }\frac{x^2}{1-\cos x}=\lim _{x\to 0 }\frac{2x}{\sin x}=\lim _{x\to 0 }\frac{2}{\cos x}=\frac{2}{\cos(0)}=1}\)

By, LHS twice.