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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?

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

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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.

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  • $\begingroup$ How about in this case?$f(x)<g(x)$,then can we write $\lim_{x \to k} f(x) < \lim_{x \to k} g(x)$? $\endgroup$
    – a_i_r
    Nov 11, 2020 at 8:14
  • $\begingroup$ $\lim f\color{red}\le\lim g$ $\endgroup$ Nov 11, 2020 at 8:15
  • $\begingroup$ @AritraBarua You can write that, but it is not true. What is true is that if the two limits exist, then $\lim_{x\to k} f(x)\leq \lim_{x\to k} g(x)$. $\endgroup$
    – 5xum
    Nov 11, 2020 at 8:15
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    $\begingroup$ @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. $\endgroup$
    – 5xum
    Nov 11, 2020 at 8:17
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    $\begingroup$ @AritraBarua I strongly suggest you also read a textbook or your lecture notes or something similar. $\endgroup$
    – 5xum
    Nov 11, 2020 at 8:20
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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.

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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 $\sin x\sim x$.

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