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Show that $$\lim_{x\to 0}\frac{\sin(\frac{x}{2})}{\frac{x}{2}}=1$$ Now I understand that we can change this limit, by computing $z=\frac{x}{2}$, so that the function becomes $$\frac{\sin(z)}{z}$$ but here's where I'm confused: Since $z=\frac{x}{2}$ this means that $x=2z$ and so I thought we'd have $$\lim_{2z\to 0}\frac{\sin(z)}{z}$$ however I know that the correct answer is $$\lim_{z\to 0}\frac{\sin(z)}{z}$$ which we know is equal to 1.

So, I'm just writing here to ask why it is $z\to 0$ instead of $2z\to 0$.

Many thanks in advance.

Luke
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    As $x\to 0$, $z=\frac{x}{2}$ tends to $0$ as well. In other words, when $2z$ tends to $0$, so does $z$. – Ayoub Apr 14 '24 at 11:55
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    If you are not sure, just do it using the definition of a limit. Formal proof (unlike intuition) will not confuse you. – Mark Apr 14 '24 at 11:56

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