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The proof of $\lim\limits_{x \to 0}\dfrac{\sin x}{x} = 1$ I remember says that because $\cos x \leq \dfrac{\sin x}{x} \leq 1$ for all $-\pi/2< x< \pi/2$ and both $\cos x$ and $1$ is going to $1$ as $x$ goes to $0$, $\frac{\sin x}{x}$ must also be going to $1$.

But if you replace $x$ with $1/w$ for all $w$ in $(-2/\pi, 2/\pi)$. But the problem is $\frac{\cos (1)}{0}$ is not a definite value. I don't know how I can use Squeezing Theorem here, help!

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4 Answers

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HINT:

$$\frac{\sin(1/x)}{1/x}=x\sin(1/x)$$

And the sine function is bounded by $1$.

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Looks complicate !

$$\left|\frac{\sin\left(\frac{1}{x}\right)}{\frac{1}{x}}\right|=\left|x\sin\left(\frac{1}{x}\right)\right|\leq |x|\underset{x\to 0}{\longrightarrow } 0.$$

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$$\frac{\sin(1/x)}{1/x}=x\sin(1/x)$$

The answer is zero because it is "zero$\times$ bounded "=$0$

$x\to 0$ and

$$-1\leq \sin (1/x) \leq 1$$

$\sin(1/x)$ is bounded

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A zero-bounded limit is one in which the function can be broken into a product of two functions where one function converges to zero and the other function is bounded. If we show that a limit is zero -bounded, then the zero-bounded limit theorem implies that the limit goes to zero.

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Instead of looking at $\frac{\sin(1/x)}{1/x}$ for small $x$, you can look at $\frac{\sin(y)}{y}$ for large $y$. Now things should be more obvious: the denominator gets large, while the absolute value of the numerator stays less then or equal to $1$. Hence, the fraction will tend to $0$.

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