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I was reviewing my MV Calculus (got a bit rusty) and was stuck at this:

Question 1

If I have a function (curve) in $N$ dimensional space which is parametricized as $g(t)$. Now, $g'(t)|_{t=a}$ gives the tangent vector at $t=a$.

The book states (without proof) that the tangent line to the curve is given by the set $$\{g(a) + t\times g'(a)\ | \text{ }t \in \mathbb{R} \}$$

How is this true? Isn't the tangent vector simply $g'(t)|_{t=a}$

Consider g(t) = $\begin{pmatrix} sin(t)\\ cos(t)\\ t \end{pmatrix}$, g'(t) = $\begin{pmatrix} cos(t)\\ -sin(t)\\ 1 \end{pmatrix}$

At $t=\frac{\pi}{2}$, g'(t) = $\begin{pmatrix} 0\\ -1\\ 1 \end{pmatrix}$

What ahead?

Question 2

Is $\lim_{x \to a} a^{f(x)}$ the same as $a^{\lim_{x \to a} {f(x)}}$ ?

a is a constant

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Ad Q1: The tangent to the curve at the point $g(a)$ is a line. The easiest way to present a line $\ell$ in space is via a parametric representation $t\mapsto{\bf x}(t)$. For this you need an initial point ${\bf p}\in\ell$ and a vector ${\bf u} \ne {\bf 0}$ pointing in the direction of $\ell$. The parametric representation of $\ell$ is then given by $$t\mapsto {\bf x}(t):={\bf p} + t\ {\bf u}\qquad(-\infty < t < \infty)\ .$$ In your case you can take ${\bf p}={\bf g}(a)$ and ${\bf u}:={\bf g}'(a)$; so the representation of the tangent becomes $$t\mapsto {\bf x}(t):={\bf g}(a) + t\ {\bf g}'(a)\qquad(-\infty < t < \infty)\ ,$$ as stated in your book.

Ad Q2: If $a$ is a positive constant then the function $y\mapsto a^y:=e^{y\log a}$ is continuous on all of ${\mathbb R}$. Assume that $\lim_{x\to a} f(x)=\eta\in{\mathbb R}$. Define $\tilde f$ by $\tilde f(a):=\eta$ and $\ :=f(x)$ otherwise. Then $\tilde f$ is continuous at $a$, and as the composition of continuous functions is continuous we have $$\lim_{x\to a}a^{f(x)}=\lim_{x\to a}a^{\tilde f(x)}=a^\eta=a^{\lim_{x\to a}f(x)}$$ (note that $f(a)$ is not inspected in a limiting process $x\to a$).

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