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I'm following a well known Calculus text book to restudy Calculus. One of the example problems in the book asks the following:

Using $$\lim_{\theta\to 0}\frac{\sin\theta }{\theta}=1,$$ show that $$\lim_{h\to 0}\frac{\cos h-1}{h}=0$$

The answer states that:

Using the formula $\cos h=1-2\sin^2(h/2)$ and $\lim_{h\to 0}\frac{\cos h-1 }{h}=\lim_{h\to 0}-\frac{2\sin^2(h/2)}{h}$

Let $\theta=h/2$,

$$\lim_{h\to 0}-\frac{2sin^2(h/2) }{h}=-\lim_{\theta\to 0}\frac{\sin(\theta) }{\theta}\sin\theta=-(1)\cdot(0)=0$$

I understand the logic behind this explanation. However, what I don't understand is: how is it possible that the solution claims that:

$$-\lim_{2\theta\to 0}\frac{\sin(\theta) }{\theta}\sin\theta=-\lim_{\theta\to 0}\frac{\sin(\theta) }{\theta}\sin\theta \ ?$$

I'd be glad if you could help me with this question. Thank you very much in advance!!!

Sebastiano
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charlietan84
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1 Answers1

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If $2 \theta$ goes to zero, $\theta$ will also go to zero.

Edit: further explanation requested in comments Hint: 2theta is continous at 0. More technical explanation ( also dont wanna be closed as duplicate) Formal basis for variable substitution in limits

  • It does provide a rough answer but should be fleshed out into a full explanation. – Jam Jul 02 '20 at 10:58
  • I will explain , what more would you like to know? –  Jul 02 '20 at 11:12
  • Imo, you should attempt to give more detail about why $\theta$ and $2\theta$ can be exchanged in the expression, the fact that multiplication by $2$ is continuous, etc. Basically just write a short paragraph about how it works. The question I'd linked in the comment under the OP question gives more technical details and justification. – Jam Jul 02 '20 at 11:23
  • Thank you , I have provided hint and further instruction. Is it ok now or should I add something more? –  Jul 02 '20 at 12:16