Twisting with limits

Question

We know,$\lim_{x \to 0} \frac{\sin x}{x}=1$,but can we write $\lim_{x \to 0} \sin x=x$? Another question,can we apply limits on both sides of an inequality?For example when we prove $\lim_{x \to 0}\frac{\sin x}{x}=1$,we deduce it from $\cos x \le \frac{\sin x}{x} \le 1$ and then apply limits on all the inequalities.Is it justified?If so,then what's the proof?

Answer 1

We know,$\lim_{x \to 0} \frac{\sin x}{x}=1$,but can we write $\lim_{x \to 0} \sin x=x$

No. On the right hand side, the expression $x$ is undefined, so the equation $\lim_{x\to 0}\sin x = x$ is meaningless.

Technically speaking, the expression is correct so long as the value of $x$ is $0$. However, in that case, the equality $\lim_{x\to 0} \sin x = \sin x$ is also technically correct.

For all practical purposes, the expression is meaningless, and dangerously so, because you use the same symbol, $x$, for two different purposes. On the left hand side, $x$ is a bound variable (meaning you cannot replace it with a fixed value and still get a valid expression), while on the right hand side, it is unbound (meaning you can replace it with a fixed value and get a valid expression, one that can be either correct or incorrect depending on what you replace $x$ with). Note the difference between a valid expression and a correct expression.

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As for your second question, the conclusion is justified because we apply a well known theorem called the Squeeze theorem.

Answer 2

Note that $\lim_{x \rightarrow 0} \sin(x) = 0$, so the equality "$\lim_{x \rightarrow 0} \sin(x)=x$" has the same meaning as "$0=x$".

When you learn about asymptotic expansions, you will be able to write something like $\sin(x)=x+o(1)$, which is the rigorous way of write what you want.

Answer 3

You're right about how we deduce the first limit. But limits, where they exist, are constants. In particular, $\lim_{x\to0}\sin x=0$. You may be more interested in the asymptotic notation $\sin x\sim x$.