Ambiguity in limits

Question

What is

$\displaystyle\lim_{x\to\frac{\pi}{2}}\bigl(\sqrt{\cos^2x-2\sin x+2\sin^2x}\bigr)^{[\sin x]}$

(where $[\,{.}\,]$ denotes greatest integer function) equal to?

I substituted the values directly and that gave me $ 0^1$. This is not an indefinite value and thus the answer should be correct. But the correct answer is $1$. How is that possible?

Answer 1

Hint

If $x\in [0,\pi]$, $x\neq \frac{\pi}{2}$ $$\left(\sqrt{\cos^2(x)-2\sin(x)+2\sin^2(x)}\right)^{[\sin(x)]}=1.$$

Answer 2

Note that $$\lim_{x\to\frac{\pi^+}{2}}\sin(x)=\lim_{x\to\frac{\pi^+}{2}}\sin(x)=1^{-}$$

$$[\sin(x)]=0$$

The expression under root $=0$ (Note that this is not $0$, but $0^+$.) $$\therefore(\text{something greater than zero})^{0/2}=1$$

In fact, as pointed out by @Surb, this holds good for any $x\in [0,\pi]$, $x\neq \frac{\pi}{2}$ since it again becomes $$(\text{something greater than zero})^{0/2}=1$$

Answer 3

Let's observe that $$ \cos^2 x - 2 \sin x + 2\sin^2 x = \sin^2 x -2\sin x + 1 = (\sin x-1)^2 $$ Then $\sqrt{(\sin x - 1)^2} = |\sin(x) - 1| = 1 - \sin x = f(x)$ [let]

Now, we have that \begin{align*} [\sin x ] = 0 \quad \text{for any } x \in \left[0, \tfrac{\pi}{2}\right) \end{align*} Hence, as $x \to \pi/2$, $f(x)^0 = 1$.

Answer 4

The greatest integer function $\lfloor\,{.}\,\rfloor$ is not continuous at $1$, and neither is the composition $f(x)=\lfloor\sin x\rfloor$ at $\pi/2$, so you're not allowed to do direct substitution.

Fix $\delta$ such that $0<\delta<\pi/2$. Then, for $x\in(\pi/2-\delta,\pi/2+\delta)$, we have $$ f(x)=\begin{cases} 0 & x\ne\pi/2 \\ 1 & x=\pi/2 \end{cases} $$ so you see that $\lim_{x\to\pi/2}f(x)=0\ne 1=f(\pi/2)$.

Set, for simplicity, $I=(\pi/2-\delta,\pi/2+\delta)$.

The function you're given is well defined in $I\setminus\{\pi/2\}$ because $\cos^2x-2\sin x+2\sin^2x=(1-\sin x)^2$ and this is nonzero for $x\in I\setminus\{\pi/2\}$. On the other hand, $\lfloor\sin x\rfloor=0$ for $x\in I\setminus\{\pi/2\}$ and therefore $$ \bigl(\textstyle\sqrt{\cos^2x-2\sin x+2\sin^2x}\,\bigr)^{\lfloor\sin x\rfloor}= \bigl(\sqrt{\cos^2x-2\sin x+2\sin^2x}\,\bigr)^0=1 $$ for $x\in I\setminus\{\pi/2\}$. This means that the limit is $1$.