What conceptual error am I making in limit evaluation?

Question

Evaluate $\lim_{x\to 0}\dfrac{\sin[\cos x]}{1+\cos x}$ where $[.]$ denotes the floor function.

Attempt:

As $x\to 0$, $\cos x \to 1$ and therefore $[\cos x]\to 1$ $\implies \lim_{x\to 0}\dfrac{\sin[\cos x]}{1+\cos x} = \dfrac{\sin 1}{2}$

However, the answer is $0$. What is the concept mistake in my solution?

Answer 1

When $x\ne0$ but $|x|<\pi/2$, then $0<\cos x<1$ so that $[\cos x]=0$ and $\sin[\cos x]=0$. Therefore $\lim_{x\to0}\sin[\cos x]=0$ etc.

Answer 2

$\lim_{x\to a} f(x) = f(\lim_{x\to a} x)=f(a)$ only if $f$ is continuous. That's pretty much the entire point of limits; that $\lim_{x\to a} f(x) \ne f(a)$.

for $x \in [0, 2\pi)$, $[\cos x] = 1$ if AND ONLY IF $x = 0$. When $x \to 0$, $x$ itself does NOT equal zero, and when $x \to 0$, $[\cos x] = 0$ and $x\to 0\not\implies [\cos x]\to [\cos 0]=1$. In fact. $x\to 0\implies [\cos x]=0\to 0$ instead.

This is the exact same reason if $f(x) = x; x < 2$ and $f(x)=5 + x^2; x \ge 2$ then $\lim_{x\to 2^-} f(x) \ne f(\lim_{x\to 2^-} x) = f(2) = 9$.

Answer 3

\begin{align} \lim_{x\to{0}^{+}} cosx = {1}^{-} \end{align} So \begin{align} \lfloor{ cosx }\rfloor = \lfloor{ {1}^{-} }\rfloor = 0 \end{align} And \begin{align} \lim_{x\to{0}^{-}} cosx = {1}^{-} \end{align} So \begin{align} \lfloor{ cosx }\rfloor = \lfloor{ {1}^{-}}\rfloor = 0 \end{align} Now you know where you were wrong