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Let $\gamma : \mathbb{R} \to \mathbb{R} ^n$ a smooth curve.

Let $\{ T_r \}_{r=1}^ \infty$ a sequence such that $T_r$ converges to a real number $L$.

Why it is true that $$\lim_{r \to \infty} \gamma (t+T_r)= \gamma (t+L)?$$ The book that I am reading says that: "Because $ \gamma$ is continuous", but I can't explain why it happens because $\gamma$ is continuous...

Can someone please explain me this?

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  • $\begingroup$ Continuity assures that $$\lim_{r\to +\infty} \gamma (t + T_r) = \gamma \left(\lim_{r\to +\infty} t + T_r \right) = \gamma (t+L)\;.$$ $\endgroup$ – Pacciu May 25 at 17:28
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Let $f: \mathbb{R} \to \mathbb{R}$ be continuous function. Now from the $\epsilon-\delta$ definition we know that for every $\epsilon > 0$ we can pick a $\delta > 0$ such that whenever $|x-y| < \delta$, $|f(x) - f(y)| < \epsilon$.

Now we are given a convergent sequence $\{x_n\}$ with limit lets say $x$. We need to show that:

$$\lim_{n \to \infty} f(x_n) = f(x)$$

Let $\epsilon > 0$, we need to find a $\delta > 0$ such that whenever $|x_n-x| < \delta$, $|f(x_n) - f(x)| < \epsilon$. So what do we do? We simply choose $\delta = \epsilon$.

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When $ \gamma (t)= (\gamma_i(t))$, then $\gamma$ is continuous iff each component $\gamma_i$ is continuous.

Hence when $\gamma$ is continuous, then $$\lim_r\ \gamma_i(t+T_r) = \gamma_i (t+L) $$ so that so is $ \gamma $

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