Twisting with limits

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?

• I think yours first question is wrong leftside be value but right side a variable Nov 11, 2020 at 8:10

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.

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

• How about in this case?$f(x)<g(x)$,then can we write $\lim_{x \to k} f(x) < \lim_{x \to k} g(x)$? Nov 11, 2020 at 8:14
• $\lim f\color{red}\le\lim g$ Nov 11, 2020 at 8:15
• @AritraBarua You can write that, but it is not true. What is true is that if the two limits exist, then $\lim_{x\to k} f(x)\leq \lim_{x\to k} g(x)$.
– 5xum
Nov 11, 2020 at 8:15
• @AritraBarua The proof of this statement can be found in any introductory calculus textbook. This site is not meant for you to list a series of questions, but rather to ask one question, and once you get the answer, you accept it and move on. Don't expect a full calculus 1 lesson as an answer to one question you posed.
– 5xum
Nov 11, 2020 at 8:17
• @AritraBarua I strongly suggest you also read a textbook or your lecture notes or something similar.
– 5xum
Nov 11, 2020 at 8:20

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.

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 $$\sin x\sim x$$.