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I've solved this problem, but I'm unsure if my reasoning is correct. Please review my understanding of the problem and whether or not my reasoning is correct. Thank you.

We know that $\lim\limits_{\theta \to 0} \frac{\sin(\theta)}{\theta} = 1$.

We want to know $\lim\limits_{t \to 0} \frac{\sin(kt)}{t}$.

$\lim\limits_{t \to 0} \frac{\sin(kt)}{t}$ = (k) $\lim\limits_{t \to 0} \frac{\sin(kt)}{kt}$. This is because we are multiplying both the numerator and denominator by $k$. Therefore, by the limit laws, we are not changing the limit?

As $t$ goes to $0$, $kt$ goes to $0$:

(k) $\lim\limits_{t \to 0} \frac{\sin(kt)}{kt}$ = (k) $\lim\limits_{kt \to 0} \frac{\sin(kt)}{kt}$.

Let $kt = \theta$.

(k) $\lim\limits_{kt \to 0} \frac{\sin(kt)}{kt}$ = (k) $\lim\limits_{\theta \to 0} \frac{\sin(\theta)}{\theta}$ = (1)k = k

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    $\begingroup$ Please use MathJax to write the math: meta.math.stackexchange.com/questions/5020/… $\endgroup$ Sep 11, 2016 at 19:02
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    $\begingroup$ This is correct. $\endgroup$ Sep 11, 2016 at 19:03
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    $\begingroup$ @Lovsovs I managed to fix it. Not sure how to include the thetas, though. :S $\endgroup$ Sep 11, 2016 at 19:12
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    $\begingroup$ For greek letters, you write \greekletter. $\endgroup$ Sep 11, 2016 at 19:22
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    $\begingroup$ @DavidBowman Thanks! $\endgroup$ Sep 11, 2016 at 19:22

2 Answers 2

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Let $t = k\theta$ then if $\theta \to 0, t \to 0$ and so;

$$\lim_{\theta \to 0} \frac{\sin(k\theta)}{\theta} = \lim_{\theta \to 0} \frac{k}{k}\cdot \frac{\sin(k\theta)}{\theta} = k \cdot \lim_{\theta \to 0} \frac{\sin(k\theta)}{k\theta} = k \cdot \lim_{t \to 0} \frac{\sin (t)}{t} =k \cdot 1 = k$$

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  • $\begingroup$ There is an error in your calculations? $\endgroup$ Sep 11, 2016 at 19:09
  • $\begingroup$ Ok, it's fixed. Everything you did was basically correct, I just submitted this as an answer to close the question. You might want to look at your substitution though, that's where you error is. $\endgroup$ Sep 11, 2016 at 19:10
  • $\begingroup$ Ok, thanks. Which substitution are you referring to? $\endgroup$ Sep 11, 2016 at 19:19
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Set $kt = x$

Hence your limit becomes

$$\lim_{x \to 0} \frac{\sin x}{x} = 1$$

Hence your limit is then $k$

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  • $\begingroup$ Please elaborate on your answer. Specify exactly where my reasoning is incorrect. $\endgroup$ Sep 11, 2016 at 19:25
  • $\begingroup$ @ThePointer It's not, sorry. You made a mess with not being able to write properly in LaTeX so I got confused. $\endgroup$
    – Enrico M.
    Sep 11, 2016 at 19:27
  • $\begingroup$ My apologies. :( $\endgroup$ Sep 11, 2016 at 19:27

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