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So I've written a complete proof for this statement, and need confirmation regarding whether or not all I've written is true.

First things first, I made the substitution $\displaystyle\tan\theta=\frac{\sin\theta}{\cos\theta}$, and from here I set up the following circle with center $O$ and radius $1$ arbitrary unit to find another limit: $\displaystyle \lim_{\theta \to 0}\frac{\sin\theta}{\theta}$ (which I will later reveal why)

enter image description here

Given this, it's evident that:

$\Delta OBA \le$ sector area $OBA \le \Delta OTA$, which is equal to $\displaystyle\frac{1}{2}\cdot1\cdot1\sin\theta \le \frac{1}{2}\cdot1^2\cdot\theta\le\frac{1}{2}\cdot1\cdot\tan\theta$

which simplifies down to:

$\displaystyle \sin\theta \le \theta \le \frac{\sin\theta}{\cos\theta}$

and from here, dividing through by $\sin\theta$ produces:

$\displaystyle 1 \le \frac{\theta}{\sin\theta} \le \frac{1}{\cos\theta}$

and since the limit of $\cos\theta$ approaches $1$ as $\theta$ approaches $0$, the following is true:

$\displaystyle 1 \le \frac{\sin\theta}{\theta} \le 1$

which implies that $\displaystyle \lim_{\theta \to 0}\frac{\sin\theta}{\theta} = 1$

And the rest is as follows:

$\displaystyle \lim_{\theta \to 0}\frac{\tan\theta}{\theta} =\lim_{\theta \to 0}\frac{\sin\theta}{\theta}\cdot \lim_{\theta \to 0} \frac{1}{\cos\theta} = 1$

therefore; $\displaystyle \lim \atop \theta \to 0$ $\displaystyle \frac{\tan\theta}{\theta} = 1$, QED

(Is all of what I've written correct?)

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    $\begingroup$ At some point you've claimed the following : $1<1$ is true (which certainly is not). Although I'm aware of what you meant, this is plainly wrong and shouldn't be written this way. When you are passing to the limits strict inequalities become non-strict ones. (and you also need to put $\lim_{\theta \to 0}$). Also, $\lim_{\theta \to 0} \theta=0$, so you are dividing by $0$. You cannot conveniently distribute limits over division/multiplication wtihout appropriate justification and conveniently pass only a part of limit to its acutal value and pull out the limit (as you seem to be doing) $\endgroup$ – user160738 Aug 21 '17 at 21:33
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    $\begingroup$ For the correct approach: $$\lim_{\theta\to 0} \frac{\tan \theta}{\theta}=\lim_{\theta\to 0}\frac{\sin \theta}{\theta} \cdot \lim_{\theta\to 0} \frac{1}{\cos \theta}$$. Both limits exist and equal to $1$, so doing this is justified ( and original limit equal to product of these limits) $\endgroup$ – user160738 Aug 21 '17 at 21:38
  • $\begingroup$ The convenient passing of limits are according to existing limit properties, such as the Quotient property, which states that $\displaystyle \lim \frac{u}{v} = \frac{\lim(u)}{\lim(v)}$, and the Product property, which states that $\displaystyle \lim uv = \lim(u)\cdot\lim(v)$ and these properties also work in reverse, but your approach does make more sense, so thank you for the feedback. :) $\endgroup$ – joshuaheckroodt Aug 21 '17 at 21:38
  • $\begingroup$ Write $\cos(\theta) \le \frac{\sin(\theta)}{\theta} \le 1$ instead of $1 < \frac{\sin(\theta)}{\theta} < 1$. $\endgroup$ – Math Lover Aug 21 '17 at 21:42
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A couple of remarks.

1- Change $<$ to $\le$.

2- Your proof presumes $\theta >0$. You also need to consider $\theta < 0$ case.

3- In both cases, establish $\cos(\theta) \le \frac{\sin(\theta)}{\theta} \le 1$ and evaluate the limits.

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  • $\begingroup$ Once you've done it for $\theta>0,$ then for $\theta<0$ you can just say $\eta = -\theta > 0$ and then $$ \frac{\tan\theta} \theta = \frac {\tan(-\eta)}{-\eta} = \frac{-\tan\eta}{-\eta} = \frac{\tan\eta} \eta. $$ As $\theta$ approaches $0$ from below then $\eta$ approaches $0$ from above and that one was already dealt with. $\qquad$ $\endgroup$ – Michael Hardy Aug 21 '17 at 22:30
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Only one thing is wrong: don't ever have separate limits for $\sin\theta, \,\theta$ in your fraction, as that gives the indeterminate form $0/0$. The limit of $fg$ only has the obvious factorization in terms of two limits if they are finite, but $1/\theta $ diverges.

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  • $\begingroup$ So how would you rewrite the fraction with respect to the position of each limit in that case? @J.G. $\endgroup$ – joshuaheckroodt Aug 21 '17 at 21:33

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