# trigonometry limits when $x$ approaches zero

a.) $\lim_{x\to 0} x/\sin x = 1$

b.) $\lim_{x \to 0} \sin x/ x = 1$

Is a.) and b.) true?

Because if I try to apply this to $\lim_{x \to 0} \sin 17x/ x$, my answer is $17$, not $1$...

• Both a) and b) are true. Nov 2, 2017 at 8:41
• But when I tried it with sin17x/ x I got 17. Nov 2, 2017 at 8:43
• This may seem counterintuitive in some sense, but the answer is actually $17$. Nov 2, 2017 at 8:44

Product of a number and an infinitesimal is still an infinitesimal. So $$17\times 0 \to 0$$. Actually an infinitesimal means a value approaching to zero. So if you multiply something (ofc not infinity) with an infinitesimal then you will get again an infinitesimal.

Next notice that $$\frac {\sin {17x}}{x} =17\frac{\sin {17x}}{17x}$$

Thus $$\lim_{x \to 0} \sin 17x/ x=\lim_{x \to 0} 17\frac{\sin {17x}}{17x}=17$$

• Yes. That is what I did. Then 17x in numerator and denominator are cancelled out. 17(sin(1)) gives me some weird answer Nov 2, 2017 at 8:46
• Do you know why this does not equal to 1 if the above laws are true?... Nov 2, 2017 at 8:47
• @user498021 You can't cancel the $17x$ from within the $\sin$ to get $\sin(1)$. Nov 2, 2017 at 8:48
• Using infinitesimal and limits in the same context is confusing.
– user312097
Nov 2, 2017 at 9:12
• "Using infinitesimal and limits in the same context is confusing"- you're totally right Nov 2, 2017 at 9:16

a) and b) are both true

$\lim\limits_{x \to 0} \frac{\sin 17x} x=17\lim\limits_{x \to 0} \frac{\sin 17x} {17x}$ Lets operate the changee of variable $y=17x$

$y\to 0$ as $x\to 0$

$17\lim\limits_{x \to 0} \frac{\sin 17x} {17x}=17\lim\limits_{y \to 0} \frac{\sin y} {y}=17\times1=17$

• yes, I undertsnad now. thank you Nov 2, 2017 at 9:10

If you can expand the taylor series of $\sin x$ this is very easy to see $$\sin x = \frac{x}{1!}-\frac{x^3}{3!}+\frac{x^5}{5!}...\\ \implies \frac{\sin x }{x}=1-\frac{x^4}{3!}+\frac{x^4}{5!}...\\ \lim_{x\rightarrow0}\frac{\sin x }{x} = \lim_{x\rightarrow0}\left(1-\frac{x^2}{3!}+\frac{x^4}{5!}...\right)=1$$ Using this you can see that taking the inverse of this would work as well.

So $$\lim_{x\rightarrow0}\frac{\sin 17x }{x} = \lim_{x\rightarrow0}\left(17-\frac{(17x)^2}{3!}+\frac{(17x)^4}{5!}...\right)=17$$

• Noooo, he is struggling with basic concepts of limits (not that it is a bad thing) and you are giving a circular answer with taylor series. How's it of any help ?
– user312097
Nov 2, 2017 at 9:33
• I see that now. Let's leave this here so he knows why $lim \sin x/x$ is equal to 1 Nov 2, 2017 at 9:36
• Isn't this circular reasoning ? How do you taylor series of $\sin$ without know $\lim_{x\to0}\sin x/x = 1$ in the first place ?
– user312097
Nov 2, 2017 at 9:39
• You don't need the limit to derive the taylor series. Taylor is a more advanced concept but this isn't circular reasoning. Nov 2, 2017 at 9:42
• This is circular reasoning. To derive the Taylor series of $f(x)=\sin(x)$ centered at $x=0$, you must use: $$f(x)=\sum_{n=0}^{\infty} \frac{f^{(n)}(0)}{n!}x^n$$ Which requires the computation of $f'(0)$ to derive the first non-zero term. Now, how can one evaluate $\sin'(0)$? Fundamentally, we must use the definition of the derivative. Hence, this is the same as computing: $$\lim_{x\to 0} \frac{\sin{x}-\sin{0}}{x-0}=\lim_{x\to 0} \frac{\sin{x}}{x}$$ Hence, you must know what $\lim\limits_{x\to 0} \dfrac{\sin{x}}{x}$ is to know the Taylor series of $\sin{x}$ you mentioned above. Nov 2, 2017 at 23:40