Continuity of A Limit of A Function Validation I would like to ask if my argument is correct because i would like to show that my equation is an indeterminate form of $1^{\infty}$
by definition of continuity we have the following:
                $$if \qquad\lim_{x \to x_{0}} f(x)=f(x_{0})\qquad then\\
    lim_{x \to x_{0}}f(g(x))=f(\lim_{x \to x_{0}}g(x))$$
                if given that $f$ is continuous at $g$ and $\lim_{x \to x_0}g(x)=g(x_{0})$
                $$\implies\qquad\lim_{t \to 0}(\frac{\sin{t}}{t})^{1/t}\qquad=\qquad(\lim_{t \to 0}\frac{\sin{t}}{t})^{1/t}$$
However, I think the problem is when I have put the limit inside the function, but I cannot limit $f(x)$ itself. How can I fix this argument so the limit can be in the form $1^\infty$?
 A: The correct result is the following:

If $\lim_{x \to x_{0}}g(x) = L$ and $f$ is continuous at $L$ then $$\lim_{x \to x_{0}}f(g(x)) = f\left(\lim_{x \to x_{0}}g(x)\right)$$

Your example is the expression $$\left(\frac{\sin t}{t}\right)^{1/t} = f(g(t))$$ so that $g(t) = \dfrac{\sin t}{t}$ and then what is $f(t)$?? You will see that there is no corresponding $f$ available from the expression given above.
The right approach is to note that $$\left(\frac{\sin t}{t}\right)^{1/t} = \exp\left(\frac{1}{t}\log\left(\frac{\sin t}{t}\right)\right) = f(g(t))$$ and then $$f(t) = \exp(t), g(t) = \frac{1}{t}\log\left(\frac{\sin t}{t}\right)$$ Clearly we can see that $g(t) \to 0$ as $t \to 0$ and hence $f(g(t)) \to f(0) = 1$ as $t \to 0$ and the desired limit of $f(g(t))$ is $1$.
Your issue is in dealing with the general power. An expression of the form $x^{y}$ always involves two things: the base and the exponent and in this form it can not be captured as a composition of functions of single variable but rather as a function of two variables $x, y$. This is true even if $x, y$ are functions of another single variable $t$. The right approach is always to express $x^{y} = \exp(y\log x)$ and then you can see that if $x, y$ depend on one variable then we can keep $g(t) = y(t)\log x(t), f(t) = \exp(t)$ so that $x^{y} = f(g(t))$.
