# Basic question regarding limits

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).

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.

• 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. – James Groon Apr 1 '17 at 18:24
• 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. – Simon Apr 1 '17 at 18:34
• Gotcha and yep, Chris did a good job :) – James Groon Apr 1 '17 at 18:35

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 $$\left|\tan \frac{1}{x}\right| < \left|\tan \frac{\epsilon}{1+\sec \epsilon} \right| = \frac{\sin \left( \frac{\epsilon}{1+\sec \epsilon}\right) }{\cos\left( \frac{\epsilon}{1+\sec \epsilon}\right)} < \frac{\sin \left( \frac{\epsilon}{1+\sec \epsilon}\right) }{\cos\epsilon} < \frac{ \frac{\epsilon}{1+\sec \epsilon} }{\cos\epsilon} = \frac{\epsilon}{1 + \cos \epsilon} < \epsilon$$

• 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 $\tan(\pi/4)=1>\pi/4$. – Mark Viola Apr 1 '17 at 18:48
• So may i get the correct answer ? – James Groon Apr 1 '17 at 19:02
• @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? – ChargeShivers Apr 2 '17 at 1:01
• @chrisvarghese You wrote $|\tan(\epsilon)|\le \epsilon$. That is false. – Mark Viola Apr 2 '17 at 2:58
• @Dr.MV Got it now. Thanks. Not sure how I missed it twice!! Answer edited. – ChargeShivers Apr 2 '17 at 17:17

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!

• $\sqrt{1 + \frac{1}{\epsilon^2}} = \dfrac{\sqrt{1+\epsilon^2}}{\epsilon}$ is definitely a better $X(\epsilon)$ than $\dfrac{1+\sec \epsilon}{\epsilon}$. May be the 'best' of all is $X(\epsilon) = \dfrac{1}{\arctan \epsilon}$. – ChargeShivers Apr 2 '17 at 23:41

Short answer is $0$ ,but if you want " to show " recall the limit $$\lim _{ x\rightarrow \infty }{ \frac { 1 }{ x } =0 }$$

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.