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Find the limit:

$$\lim_{x\to 1^+}\left\lfloor\sqrt\frac{1-\sqrt{x-1}}{1+\sqrt{x-1}}\right\rfloor$$

where $\lfloor x\rfloor$ is the floor function.

Generally :

How can \begin{align} \lim_{x\to a^-}f(x)&=\lim_{x\to a}g(x) \\ \lim_{x\to a^+}f(x)&=\lim_{x\to a}g(x)? \end{align}

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    Hint: If the limit exists, it must be because the function value is _constant_ in some interval $(1,1+\varepsilon)$. – 2017-02-21
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    Start simplifying the function inside the floor function, by example multiply denominator and numerator by $1-\sqrt{x-1}$. Probably it will help you remember that the floor function can be divided in a set of constant functions. – 2017-02-21
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    At $x=1$, the value is clearly $1$. However, approaching from the left will give you a complex number, and approaching from the right will give you $0$ – 2017-02-21
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    Following @HenningMakholm. Replace $x$ by $1+\epsilon^2$ and look at what happens for $\epsilon\to 0^+$... – 2017-02-21
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    Can't understand your "Generally" question. – 2017-02-21

2 Answers 2

2

Using these two approximation $\sqrt{1-\varepsilon_1}\approx1-\dfrac12\varepsilon_1$ and $\dfrac{1+\varepsilon_1}{1+\varepsilon_2}\approx1+\varepsilon_1-\varepsilon_2$‎ where $\varepsilon_1$ and $\varepsilon_2$ are small numbers we see for $x=1+\varepsilon^2$ that $$\lim_{x\to 1^+}\left\lfloor\sqrt\frac{1-\sqrt{x-1}}{1+\sqrt{x-1}}\right\rfloor= \lim_{x\to 1^+}\left\lfloor\sqrt\frac{1-\varepsilon}{1+\varepsilon}\right\rfloor= \lim_{x\to 1^+}\left\lfloor\sqrt{1-2\varepsilon}\right\rfloor= \lim_{x\to 1^+}\left\lfloor1-\varepsilon\right\rfloor=0 $$

2

For $x\in(1,2)$, we have

$$0<1-\sqrt{x-1}<1$$ and $$1+\sqrt{x-1}>1$$

so that the ratio is in $[0,1)$ and the floor of its square root is $0$.