I saw this question and wondered how OP of that question was able to do : $$0<\sin x+1<2$$ this $$\frac 0{|x|}<\frac{\sin x+1}{|x|}<\frac 2{|x|}$$ and when $x\to \infty$ he got the limit evaluated as zero.

Why i wondered is because it is not working on this inequality $$2\leq x+ \frac 1x\leq 20$$ where $x\in [1,b]$ and if i multiply both sides with $|x|=x$ in the same way in which it is done in above question the i will get $$2x\leq x^2+1\leq 20x$$ and then i take limit i get $$\lim_{x\to 0}2x\leq \lim_{x\to 0} (x^2+1)\leq\lim_{x\to 0}20x$$ which implies $1=0$ and this is false!

One may say that $0$ is not in domain of $x/x$ but we are allowed to take limit at $0$.

So where is my conceptual error?


On basis of comments i will give this example where we have restricted domain an still we can apply sandwich theorem $$\frac 1{1+|x|}\leq \frac{e^x-1}{x}\leq 1+|x|(e-2)$$ this inequality is true for any value of x in $[-1,1]-{0}$ on applying limit $$\lim_{x\to 0}\frac 1{1+|x|}\leq \lim_{x\to 0}\frac{e^x-1}{x}\leq \lim_{x\to 0}(1+|x|(e-2))$$ we get correct value of expression. And in this case also $0$ is not in the range if $x$ but we are allowed to take its limit at $x=0$.

  • $\begingroup$ But the starting point $ 2\leq x+1/x\leq 20$ is NOT true for any $x>0$!!! So of course you obtain a false statement when you make $x\to 0$. $\endgroup$ – guestDiego May 22 '16 at 14:44
  • $\begingroup$ No i have $x\in [1,b]$ and that equality is true in this range $\endgroup$ – ramsay May 22 '16 at 14:46
  • 1
    $\begingroup$ Of course, it is true in that range: $[1,b]$. But the limiting point, i.e. 0 is NOT in that range! $\endgroup$ – guestDiego May 22 '16 at 14:48
  • $\begingroup$ You have not runned into a mistery of mathematics! $\endgroup$ – guestDiego May 22 '16 at 14:49
  • $\begingroup$ @guestdiego see the last lines i edited $\endgroup$ – ramsay May 22 '16 at 14:53

Here is an example that has problems akin to your example.

$$1\le x\le2x$$

For $x\in[1,2]$.

We want to evaluate $\lim_{x\to0}$.


Can you figure out what's wrong here?

  • $\begingroup$ example is great and makes sense! but the reason why sandwich is not working, i am not able to make out? $\endgroup$ – ramsay May 22 '16 at 15:47
  • 1
    $\begingroup$ @ramsay Try and pick out what parts are important when I created my inequalities, and see what fails when I do the limit. Think about it for a few minutes and tell me what you think. $\endgroup$ – Simply Beautiful Art May 22 '16 at 15:48
  • 1
    $\begingroup$ I think that inequality is not true when $x\to 0$ that inequality is only valid when $x\in [1,2]$ $\endgroup$ – ramsay May 22 '16 at 16:02
  • $\begingroup$ @ramsay And that's why your inequality failed. $\endgroup$ – Simply Beautiful Art May 22 '16 at 16:46
  • $\begingroup$ But even in case of that $(e^x-1)/x$ The inequality is true only when $x\in [-1,1]-\{0\}$ , but why do we get correct answer when we take limit at 0 even though it is not in the interval?(sorry) $\endgroup$ – ramsay May 22 '16 at 17:15

When you let $x \to 0$, you cannot claim $x \in [1,b]$ anymore.

  • $\begingroup$ But when we apply sandwich theorem we don't change the range. $\endgroup$ – ramsay May 22 '16 at 14:44
  • $\begingroup$ @ramsay No, you can't take $x\to0$ and stay in the range. At most, you can take $x\to1$, or you need to adjust the range to include the limit. $\endgroup$ – Simply Beautiful Art May 22 '16 at 14:48
  • $\begingroup$ @simpleArt but why this guys is not changing the range $\endgroup$ – ramsay May 22 '16 at 14:50
  • $\begingroup$ @ramsay Because the limit is already inside the range, so there is no need. His equality holds for $x\in\mathbb{R}$, so no worries. $\endgroup$ – Simply Beautiful Art May 22 '16 at 14:50

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for?Browse other questions tagged or ask your own question.