I am trying to calculate the limit, using L'Hospital's Rule.
$$\lim_{x \to \infty} \left( \frac{1}{\sin^2(x)} – \frac{1}{x^2} \right)$$
My attempt
$$\lim_{x \to \infty} \left( \frac{1}{\sin^2(x)} – \frac{1}{x^2} \right) = \lim_{x \to \infty} \frac{(x^2 – \sin^2(x))'}{(x^2\sin^2(x))'} = \lim_{x \to \infty} \frac{2x – \sin(2x)}{2x\sin^2(x) +x^2\sin(2x)} $$
I stopped trying at that point because the limit seems to get overly complicated and I run out of other ideas. Any tips on how to solve this?
Extra side-note question: I generally stuggle when I try to solve limits that involve infinity with trigonometrics. Is there a general rule to reduce these problems into easier ones?
Best Answer
This limit cannot exist because $\sin (\infty)$ bounded but uncertain real number. Take two sequences $x_n=n\pi, x'_n=(n+1/2)\pi$ such that bot $x_n$ and $x_n$ tend to $\infty$. As $n \to \infty$ as $f(x)=\frac{1}{\sin^2 x}$ attains two unequal values: $f(x_n)=\infty$ and $f(x'_n)=1,$ the limit does not exist.