Two different ways to determine $\lim_{x\to \infty} \frac{\sin x}{x}$ . Method 1: By using Sandwich Theorem
We know that
$$\frac{-1}{x} \leq \frac{\sin x}{x} \leq \frac{1}{x}   $$
Applying limit,
$$\lim_{x\to \infty}\frac{-1}{x} \leq \lim_{x\to \infty}\frac{\sin x}{x} \leq \lim_{x\to \infty}\frac{1}{x}   $$
$$ 0 \leq \lim_{x\to \infty}\frac{\sin x}{x} \leq 0  $$
$$\therefore\lim_{x\to \infty}\frac{\sin x}{x} = 0$$
Method 2:
Put $x = \frac{1}{t}$
Then as $x \to \infty$ , $\frac{1}{t} \to 0.$
Equation becomes:
$$\lim_{\frac{1}{t}\to 0}\frac{\sin \frac{1}{t}}{\frac{1}{t}}$$
which is 1.
Am I doing something wrong? Please help me out.
 A: You made a mistake with the substitution.
If $x=\frac1t$, then as $x\to\infty,\ \frac1t=x\to\infty$, not zero.
The correct way to solve it is to recognise that as $x\to\infty,\ \color{red}{t}\to0^+$ (there's no use in finding where $\frac1t$ goes since it is just $x$ and you wouldn't be using the substitution).
Then, $\lim_{x\to \infty} \frac{\sin x}{x}$ becomes:
$$\lim_{t\to0^+}\frac{\sin(1/t)}{(1/t)}=\lim_{t\to0^+}t\sin(1/t)$$
Applying the squeeze theorem (since $t\to0^+,\ t$ is positive and we can multiply by it):
$$-1\leq\sin(1/t)\leq1\implies -t\leq t\sin(1/t)\leq t$$
Taking the limit of all terms:
$$\lim_{t\to0^+}(-t)\leq\lim_{t\to0^+}t\sin(1/t)\leq\lim_{t\to0^+}t$$
$$0\leq\lim_{t\to0^+}t\sin(1/t)\leq0$$
Therefore
$$\lim_{t\to0^+}t\sin(1/t)=0,$$
which is equivalent to the original limit
$$\lim_{t\to0^+}\frac{\sin(1/t)}{(1/t)}=0.$$
A: The two evaluation are indeed correct but these are two completely different limits. To make them the same we should take $x=\frac 1 t \to \infty$ with $t\to 0^+$ (and not $\frac1t \to 0^+$) and then
$$\lim_{x\to \infty}\frac{\sin x}{x}= \lim_{t\to 0^+}\frac{\sin \frac1t}{\frac1t}=0$$

Edit
As an alternative to sandwich theorem, we can proceed by contradiction assuming that
$$\frac{\sin x}{x}\to L\neq 0$$
but for $x_n= 2\pi n\to \infty$ we obtain
$$\frac{\sin x_n}{x_n}=\frac{\sin\left( 2\pi n\right)}{2\pi n}= 0$$
which is a contradiction.
Therefore if limit exists it must be equal to zero and this finally can be checked by the definition.
