Difference in limits because of greatest-integer function

A Problem: $$\lim_{x\to 0} \frac{\sin x}{x}$$ causes the remedy: $1$

But the very same function confined in a best integer function causes a $0$

$$\lim_{x\to 0} \left\lfloor{\frac{\sin x }{x}}\right\rfloor$$

Why?

My ideas: ¢ [The value of the first function often tends to 1 as a result of the development :

$$\frac{\sin\left( x \right)}{x}\approx\frac{ x - \frac{x^3}{3!} + \frac{x^5}{5!} - \frac{x^7}{7!} + \cdots}{x}$$ $$\approx 1 - \frac{x^2}{3!} + \frac{x^4}{5!} - \frac{x^6}{7!} + \cdots$$ and also placing absolutely no in the function causes 1

yet using the best integer function to the very same will certainly cause an absolutely no as whenever the value of the outcome is taken it will certainly be a little much less than one as a result of all the reductions associated with the expansion.]

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2022-07-25 20:47:02
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