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What does the following limit evaluate to? $$\lim_{x\to0^-}\frac{\sin\lfloor x\rfloor}x$$ I know that $$\lim_{x\to0}\frac{\sin x}x=1$$ but how to evaluate the above given limit.

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$$\lim_{h\to0}\frac{\sin\lfloor 0-h\rfloor}{0-h}$$ where $h>0$

$$\lim_{h\to0}\frac{\sin\lfloor -h\rfloor}{-h}$$

$$\lim_{h\to0}\frac{\sin(-1)}{-h}$$ $$\lim_{h\to0}\frac{-0.8414709848}{-h}$$(for -1 radian)

Please note that it is not an indeterminate form.Hence the limit doesnot exist.


By definition

In mathematics, a limit is the value that a function or sequence "approaches" as the input or index approaches some value.

Infinity is not that some value.

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  • $\begingroup$ it's +infinite as i said before. $\endgroup$ – hamam_Abdallah Sep 28 '16 at 16:23
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if $x$ is near 0 by right, then $\lfloor x \rfloor=0$ and $f(x)=0$ so limit by right is zero (0).

if $x$ is near 0 by left, then $\lfloor x \rfloor=-1$ and $f(x)=-1/x$, so limit by left is $+\infty$.

Thus the limit at 0 doesn't exist.

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  • $\begingroup$ But I am asking limit x tending to $0^{-}$ $\endgroup$ – sai saandeep Sep 28 '16 at 15:25
  • $\begingroup$ it's sin(-1)/0^-=+ infinity. $\endgroup$ – hamam_Abdallah Sep 28 '16 at 15:26

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