# Question about continuity of a curve

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?

• Continuity assures that $$\lim_{r\to +\infty} \gamma (t + T_r) = \gamma \left(\lim_{r\to +\infty} t + T_r \right) = \gamma (t+L)\;.$$ – Pacciu May 25 at 17:28

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 $$\delta = \epsilon$$.
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 $$\gamma$$