Consider the function $f:\{x\in\mathbb{R}, x\neq k\pi, k\in\mathbb{Z}\}\to\mathbb{R}$ given by
$$f(x) =\frac{\sin^4 x -\cos x}{\sin^3 x}.$$
As an exercise of my calculus class, the Professor asked to find an upper and lower bound for that function.
Actually, the unique bounds I know for $\sin x$ and $\cos x$ are $-1\le\sin x, \cos x\le 1$, so I don't know to obtain an upper/lower bound in this case.
Moreover, for $x\sim 0$, the term $1/\sin^3 x$ tends to infinity and then I don't know how to manage it.
Could someone please help me to find the bound for this function?
Thank you in advance!
Best Answer
For any $ x $ in the domain, we can write $$f(x)=\sin(x)-\frac{\cos(x)}{\sin^3(x)}$$ $$=g(x)+h(x)$$ As you know, the function $ g $ is bounded by $ -1$ and$ 1$, But
$$\lim_{x\to 0^-}h(x)=\frac{-1}{0^-}=\infty$$ and $$\lim_{x\to 0^+}h(x)=\frac{-1}{0^+}= -\infty$$
thus, the function $ f $ is not bounded .