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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/… $\endgroup$ – Bobson Dugnutt Sep 11 '16 at 19:02
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    $\begingroup$ This is correct. $\endgroup$ – Faraad Armwood Sep 11 '16 at 19:03
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    $\begingroup$ @Lovsovs I managed to fix it. Not sure how to include the thetas, though. :S $\endgroup$ – The Pointer Sep 11 '16 at 19:12
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    $\begingroup$ For greek letters, you write \greekletter. $\endgroup$ – David Bowman Sep 11 '16 at 19:22
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    $\begingroup$ @DavidBowman Thanks! $\endgroup$ – The Pointer Sep 11 '16 at 19:22
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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? $\endgroup$ – The Pointer Sep 11 '16 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. $\endgroup$ – Faraad Armwood Sep 11 '16 at 19:10
  • $\begingroup$ Ok, thanks. Which substitution are you referring to? $\endgroup$ – The Pointer Sep 11 '16 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. $\endgroup$ – The Pointer Sep 11 '16 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. $\endgroup$ – Turing Sep 11 '16 at 19:27
  • $\begingroup$ My apologies. :( $\endgroup$ – The Pointer Sep 11 '16 at 19:27

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