Q:

How to prove that $x\mapsto x^{\frac35}$ is continuous

Trying to teach myself calc and I know what continuity is but how exactly do we write a formal proof proving that a function is continuous. 
Say the function $f(x) = x^{\frac35}$. How would I write the step-by-step proof of why it is continuous?

A:

We have $x^{\frac35}=e^{\frac35\log x}$. Since $\log$ is defined as $$\log x = \int_1^x \frac1t \ \mathsf dt$$ for $x>0$ by the fundamental theorem of calculus, $\log$ is continuous. Moreover, $e^x = \lim_{n\to\infty} \left(1+\frac xn\right)^n$. Since $$g_n(x) := \left(1+\frac xn\right)^n = \sum_{k=0}^n \binom nk \left(\frac xn\right)^k $$ is a polynomial, $g_n$ is continuous. As $$g_n'(x) = \left(1+\frac xn\right)^{n-1} + \frac1n\left(1+\frac xn\right)^n = \left(1 +\frac xn \right)^{n-1}\left(1+\frac1n\right) $$
and $g_n'(x)=0$ implies $x=-n$. Since $g_n'\geqslant 0$ on $[0,\infty)$, and $[0,\infty)$ is a connected set which is not bounded above, it follows that any extreme value must be obtained at $x=0$. Hence
$$\sup_{x\in[0,\infty)}|g_n(x) - e^x| = |g_n(0) - e^0| =0. $$
Therefore $g_n$ converges uniformly to $\exp$, from which we conclude that $\exp$ is continuous. Now, $f(x) = g(h(x))$ where $g(x) = e^x$ and $h(x)=\frac35\log x$. The composition of continuous functions is continuous, so $f$ is indeed continuous.

