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Unsure about my solution. would appreciate your feedback.

$$\lim_{x\to 0} \frac{\sqrt{\sin(\tan(x))}}{\sqrt{\tan(\sin(x))}}= \lim_{x\to 0} \frac{\sqrt{\frac{\sin(\tan(x))}{\tan(x)}\tan(x)}}{\sqrt{\frac{\tan(\sin(x))}{\sin(x)}\sin(x)}} \tag{*}$$

We know $\lim_{x\to 0} \sin(x)\to x$ and $\lim_{x\to 0} \tan(x)\to x$; then: $$\lim_{x\to 0} \frac{\sqrt{\frac{\sin(\tan(x))}{\tan(x)}\tan(x)}}{\sqrt{\frac{\tan(\sin(x))}{\sin(x)}\sin(x)}}=\lim_{x\to 0} \frac{\sqrt{\tan(x)}}{\sqrt{\sin(x)}}=\lim_{x\to 0} \sqrt{\frac{\sin(x)}{\cos(x)\sin(x)}}=\lim_{x\to 0} \sqrt{\frac{1}{\cos(x)}}=1$$

Can I do that? Is what I did at (*) legal because of the continuity of the discussed functions?

Thanks

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You're having the right idea, but the realization is not completely sound.


The function you want to compute the limit of is defined for $0<x<\pi/2$ (disregarding other intervals that are not of a concern here). The limit has no meaning for $x<0$, so somebody would consider it as the limit for $x\to 0^+$.

On the other hand, if the function is written as $$ \sqrt{\frac{\sin(\tan x)}{\tan(\sin x)}} $$ it is defined in a whole punctured neighborhood of $0$.

Now it's a matter of computing $$ \lim_{x\to0}\frac{\sin(\tan x)}{\tan(\sin x)} $$ which we'll take the square root of after computing it. Recall that $\sin t=t+o(t)$ and $\tan t=t+o(t)$, so $$ \lim_{x\to0}\frac{\sin(\tan x)}{\tan(\sin x)}= \lim_{x\to0}\frac{\tan x+o(\tan x)}{\sin x+o(\sin x)}= \lim_{x\to0}\frac{x+o(x)+o(x+o(x))}{x+o(x)+o(x+o(x))}= \lim_{x\to0}\frac{x+o(x)}{x+o(x)}=1 $$

It's not really different from your method, which is just rewriting the function as $$ \lim_{x\to0}\sqrt{\frac{\sin(\tan x)}{\tan x}\frac{\sin x}{\tan(\sin x)} \frac{\tan x}{\sin x}} $$ where each fraction under the square root sign has limit $1$.

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  • $\begingroup$ I appreciate your detailed answer, and I learnt from it few new things. but one thing I couldn't understand, and it's about the interval of definition. why isn't it relevant to look at $x<0$? why is the limit defined as for $lim_{x\to 0^+}$? $\endgroup$ – Ami Gold Jun 8 '16 at 18:00
  • $\begingroup$ @Ami For $-\pi/2<x<0$, both $\sin x$ and $\tan x$ are negative and so also $\sin(\tan x)$ and $\tan(\sin x)$ are negative. Therefore $\sqrt{\sin(\tan x)}$ and $\sqrt{\tan(\sin x)}$ are undefined. $\endgroup$ – egreg Jun 8 '16 at 18:06
  • $\begingroup$ now I feel stupid. Just noticed that.. thanks! $\endgroup$ – Ami Gold Jun 9 '16 at 6:46
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Well, what you did is OK, but if you are not capable of explaining why it's OK, then you should do more work. This is mathematics afterall, so we are only interested in things you can prove, not hunches. I can have a hunch about a great deal of things, but if I am not sure, and cannot prove them, they may as well not exist.

If I were you, I would write down

$$\lim_{x\to 0} \frac{\sqrt{\frac{\sin(\tan(x))}{\tan(x)}\tan(x)}}{\sqrt{\frac{\tan(\sin(x))}{\sin(x)}\sin(x)}}=\lim_{x\to 0} \frac{\sqrt{\tan(x)}}{\sqrt{\sin(x)}}\cdot\sqrt{\frac{\sin(\tan x)}{\tan x}}\cdot \sqrt{\frac{\sin x}{\tan(\sin x)}}$$

Then decompose the limit into a product of three limits and show each one is equal to $1$.

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Using equivalents, $\sin x\sim x$ and $\tan x\sim x$ $$\lim\limits_{x\to 0} \frac{\sqrt{\sin(\tan(x))}}{\sqrt{\tan(\sin(x))}}=\lim\limits_{x\to 0} \frac{\sqrt{\tan(x)}}{\sqrt{\sin(x)}}=\lim\limits_{x\to 0} \frac{1}{\sqrt{\cos(x)}}=1$$

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What you have done is correct without any issues (user egreg points a minor issue in your approach about left and right hand limits and you should take care of this). However not many students are aware of why your procedure is correct. While evaluating limit of complicated expressions it is possible to replace a sub-expression by its limit under certain circumstances. One of these circumstances is that the sub-expression has a non-zero limit and the sub-expression occurs in a multiplicative manner in the overall expression and this is applicable here and you have implicitly (or perhaps unknowingly, I am not sure) made use of this fact.

Another point is that the statements like $\lim_{x \to 0}\sin x \to x$ don't mean anything. No such statements are defined in the theory of limits. It appears that you are trying to say $\lim_{x \to 0}\dfrac{\sin x}{x} = 1$ and it is better to say it in this form or perhaps like $(\sin x)/x \to 1$ as $x \to 0$ or sometimes $\sin x \sim x$ as $x \to 0$, but don't write $\lim_{x \to 0}\sin x \to x$.


Here is a simpler way. First of all note that roots are here only to complicate the matters. Use the fact that $$\lim_{x\to 0}\frac{\sin \tan x}{\tan \sin x} = \lim_{x \to 0}\frac{\sin \tan x}{\tan x}\cdot\frac{\tan x}{\sin x}\cdot\frac{\sin x}{\tan \sin x} = \lim_{x \to 0}1\cdot \frac{1}{\cos x}\cdot 1 = 1$$ and then use the fact that $f(x) = \sqrt{x}$ is continuous at $x = 1$ so that $$\lim_{x \to 0}\sqrt{\frac{\sin \tan x}{\tan \sin x}} = \sqrt{1} = 1$$

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