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Find the following limit

$$\lim_{x \to 0}\frac{[x]}{x}$$

Where $[x]$ is the greatest integer function.

I tried using the squeeze theorem on this but couldn’t come up with the appropriate functions, can someone help me out?

Edit: I can now see that this limit will not exist.

But what can we say about

$$\lim_{x \to 0}x[1/x]$$

Where again [x] is the greatest integer function, how can i use squeeze theorem to find this?

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  • $\begingroup$ There's no limit. If $x>0$ is really small the term is zero. If $x<0$ is really small we get $-\frac{1}{x}$ which goes to $-\infty$. $\endgroup$ – Yanko Jan 15 '19 at 13:55
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    $\begingroup$ For your edit: Well now it seems like the limit is $1$. You need to calculate the limit for $x\rightarrow 0$ but $x>0$ and for $x\rightarrow 0$ but $x<0$. $\endgroup$ – Yanko Jan 15 '19 at 14:01
  • $\begingroup$ Are you sure you didn't mean least integer function? $\endgroup$ – Cuhrazatee Jan 15 '19 at 14:36
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if the GIF function contains a value that tends to infinity then the gif function can be removed. hence [1/x] in the second expression will simply become 1/x. now if the GIF contains some other expression then we must understand that the limit is applied on value given by GIF and not on the expression in the GIF. Right Hand Limit And Left hand limit must be found(bcoz in negative inputs it gives further negative number ie [-0.2] = -1) and then solved. Ex. $ lim_{x->0^+}[sinx/x] $ now since sine of a value is always less than the actual value hence sinx/x will be little less than 1 so GIF will return zero so = lim_{x->0}0=0

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Defining $u={1\over x}$ we must obtain two limits$$\lim_{u\to \infty}{\lfloor u\rfloor\over u}$$and $$\lim_{u\to -\infty}{\lfloor u\rfloor\over u}$$for $u\to \infty$ we can write$${\lfloor u\rfloor\over \lfloor u\rfloor+1}< {\lfloor u\rfloor\over u}\le1$$and for $u\to -\infty$$$1\le {\lfloor u\rfloor\over u}<{\lfloor u\rfloor\over \lfloor u\rfloor+1}$$by Squeeze theorem, both tend to $1$ and so does $\lim_{x\to 0}x\lfloor{1\over x}\rfloor$

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