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I'm a little confused when it comes to question like this.

Let's say we got this expression: $$\lim_{x\to\infty} \tan\left(\frac{1}{x}\right).$$ Am I allowed to say the result of this is $0$ or do I have to show it? (If I have to show it, please show me how to cuz I got no clue).

And more general, in which situations are we allowed to "cut" it and in which we are not? (I'm guessing whenever it comes to fractions surely).

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That depends.

If the whole exercise is just "compute $\lim \tan(1/x)$", then yes, you have to do some argument. Something along the lines of "if $x$ goes to infinity, then $1/x$ goes to zero, and by continuity of $\tan$, the whole expression then goes to zero".

But if this limit is just one part of a longer question/calculation, then no. It's trivial enough that anybody with some math background will see it immediately.

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  • $\begingroup$ Appreciated . Got only one more question . Is there any like more formal ways of prooving it is 0 ? Cuz I never did any good solving math problem using words tho I completely see where ur going at. $\endgroup$ – James Groon Apr 1 '17 at 18:24
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    $\begingroup$ In general feel free to replace the word "then" or the phrase "it then follows" with the symbol "$\implies$" and "there is an $x$" with "$\exists x$" and such. Using such symbols instead of words can make a proof more concise, though overdoing it can make it hard to read. And if you want a more elementary proof, the answer by @Chris-Varghese is quite nice. $\endgroup$ – Simon Apr 1 '17 at 18:34
  • $\begingroup$ Gotcha and yep, Chris did a good job :) $\endgroup$ – James Groon Apr 1 '17 at 18:35
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You need to show that given any small $\epsilon > 0$, there exists a $X(\epsilon)$ such that $|\tan(1/x)| < \epsilon,\; \forall x > X(\epsilon) $. To do this choose, for e.g., $X(\epsilon) = \dfrac{1+\sec \epsilon}{\epsilon}$. Then $$\left|\tan \frac{1}{x}\right| < \left|\tan \frac{\epsilon}{1+\sec \epsilon} \right| = \frac{\sin \left( \frac{\epsilon}{1+\sec \epsilon}\right) }{\cos\left( \frac{\epsilon}{1+\sec \epsilon}\right)} < \frac{\sin \left( \frac{\epsilon}{1+\sec \epsilon}\right) }{\cos\epsilon} < \frac{ \frac{\epsilon}{1+\sec \epsilon} }{\cos\epsilon} = \frac{\epsilon}{1 + \cos \epsilon} < \epsilon$$

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  • $\begingroup$ The inequality is wrong. Note that $|\tan (x)|\ge |x|$ for all $x\in (-\pi/2,\pi/2)$. Simple example; $x=\pi/4<1$. We have $\tan(\pi/4)=1>\pi/4$. $\endgroup$ – Mark Viola Apr 1 '17 at 18:48
  • $\begingroup$ So may i get the correct answer ? $\endgroup$ – James Groon Apr 1 '17 at 19:02
  • $\begingroup$ @Dr.MV Which inequality in my answer are you referring to? Indeed, $|\tan x | \geq |x| \forall x \in (-\pi/2, \pi/2)$. How does that fact contradict anything that I have in my answer? $\endgroup$ – ChargeShivers Apr 2 '17 at 1:01
  • $\begingroup$ @chrisvarghese You wrote $|\tan(\epsilon)|\le \epsilon$. That is false. $\endgroup$ – Mark Viola Apr 2 '17 at 2:58
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    $\begingroup$ @Dr.MV Got it now. Thanks. Not sure how I missed it twice!! Answer edited. $\endgroup$ – ChargeShivers Apr 2 '17 at 17:17
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The answer is that it depends on the situation. To address the statement in the OP "If I have to show it, please show me how to cuz I got no clue," I thought it might be instructive to present an approach that relies on a standard inequality from elementary geometry. To that end, we begin with a short primer.


PRIMER:

Recall from elementary geometry the inequality

$$\sin(\theta)\le \theta\tag 1$$

for $\theta\ge 0$. Squaring both sides of $(1)$, using $\sin^2(\theta)=1-\cos^2(\theta)$, and rearranging, we find that

$$\cos(\theta)\ge \sqrt{1-\theta^2} \tag 2$$

for $0\le \theta \le 1$.


Now, using $(1)$ and $(2)$ with $\theta=1/x$ reveals that

$$\begin{align} \tan(1/x)&=\frac{\sin(1/x)}{\cos(1/x)}\\\\ &\le \frac{1/x}{\sqrt{1-\frac1{x^2}}}\\\\ &=\frac{1}{\sqrt{x^2-1}}\tag 3 \end{align}$$

for $x>1$.


Hence, for all $\epsilon>0$, we have from $(3)$ that

$$\begin{align} \tan(1/x)&\le \frac{1}{\sqrt{x^2-1}}\\\\ &<\epsilon \end{align}$$

whenever $x>\sqrt{1+\frac{1}{\epsilon^2}}$. And we are done!

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  • $\begingroup$ $\sqrt{1 + \frac{1}{\epsilon^2}} = \dfrac{\sqrt{1+\epsilon^2}}{\epsilon}$ is definitely a better $X(\epsilon)$ than $ \dfrac{1+\sec \epsilon}{\epsilon}$. May be the 'best' of all is $X(\epsilon) = \dfrac{1}{\arctan \epsilon}$. $\endgroup$ – ChargeShivers Apr 2 '17 at 23:41
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Short answer is $0$ ,but if you want " to show " recall the limit $$\lim _{ x\rightarrow \infty }{ \frac { 1 }{ x } =0 } $$

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You can move the limit into the $tan$, because you are not excluding any $x$ from the limit, and you are not creating any limits that don't exist. Once that is done, the problem is trivial.

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