Question:
If $$\lim_{x \to 0}{\frac{-1 + \sqrt{(\tan x - \sin x) + \sqrt{(\tan x - \sin x) + \sqrt{(\tan x - \sin x) + \cdots \infty}}}}{-1 + \sqrt{x^3 + \sqrt{x^3 + \sqrt{x^3 + \cdots \infty}}}}} = \frac{1}{k}$$
Then find the value of $k$.
The way I approached the problem was by substituting $x = 0$ in the limit:
$$\frac{-1 + \sqrt{(\tan 0 - \sin 0) + \sqrt{(\tan 0 - \sin 0) + \sqrt{(\tan 0 - \sin 0) + \cdots \infty}}}}{-1 + \sqrt{0^3 + \sqrt{0^3 + \sqrt{0^3 + \cdots \infty}}}} = \frac{1}{k}$$
$$\implies \frac{-1 + \sqrt{0 + \sqrt{0 + \sqrt{0 + \cdots \infty}}}}{-1 + \sqrt{0 + \sqrt{0 + \sqrt{0 + \cdots \infty}}}} = \frac{1}{k}$$
$$\implies \frac{-1}{-1} = \frac{1}{k}$$
$$\implies \frac{1}{1} = \frac{1}{k}$$
$$\implies k = 1$$
But according to the given solution, the answer is 2.
I did find this question, but I do not understand why I cannot just put $x = 0$ in the limit.
Any help is appreciated.
Okay. when we look at these continued roots....lets look at the one in the numerator, the logic will hold for the one in the denominator.
First the roots will not be defined as real functions if $x<0$
$\lim_\limits{x\to 0} \tan x - \sin x = 0$
$\tan x - \sin x > 0$ for all $0<x<\frac {\pi}2$
When x is small and positive then $\tan x - \sin x$ will also be small and positive.
$\tan x - \sin x<\sqrt{\tan x - \sin x}$
We are looking at an infinite sequence.
$s_{n+1} = \tan x - \sin x + \sqrt {s_{n}}$
Where we are taking small positive numbers and adding something on, and since this sequence is infinite, it will eventually be something of non trivial magnitude no matter how small $\tan x - \sin x$ may have been at the start.
Eventually, it approaches 1, but it can't get more than trivally larger than one.
Because if $s_n > 1, 1<\sqrt {s_n} < s_n$
This is why you can't treat it like you do continuous functions and just plug zero.