The first statement follows from the rules of calculation for the sum, the product and the quotient of two limits. For example, for the sum, we have to show that #\lim_{x\to a} (f+g)(x) = (f+g)(a)#. This follows from the following steps, in which we use the definition of the sum of two functions, a rule of calculation for limits, the continuity of #f# and #g# in #a# and once again the definition of the sum of two functions #g# in #a#.

\[\begin{array}{rcl}\lim_{x\to a} (f+g)(x) &=& \lim_{x\to a} (f(x)+g(x))\\ &=& \lim_{x\to a} f(x)+ \lim_{x\to a} g(x) \\ &=& f(a)+ g(a)\\ &=& (f+g)(a)\end{array}\]

The second statement we prove using Limits of continuous functions. We write #b=f(a)#. Because #f# is continuous in #a#, we have #\lim_{x\to a}f(x) = b#. According to the theory given, we have #\lim_{x\to a}g(f(x)) = g(b)#. But this can be rewritten as #\lim_{x\to a}g\circ f(x) = g\circ f(a)#. This is the definition of continuity of #g\circ f# in #a#.

This follows from the statement: the function #x^d# is continuous on #\ivoo{0}{\infty}#.

Indeed the original statement follows, by applying rule 2, the continuity of the composition of continuous functions, with #g(x)=x^d#.

For a complete proof, we must now prove the statement that #x^d# is continuous. We will only do this only in the case #d# is rational.

When #d# is an integer #n\in\mathbb{Z}\backslash\{0\}#, then the statement follows from applying rule 1: the product of two or more continuous functions is again a continuous function. The rule speaks of the product of two functions, but by applying the rule repeatedly, we see that it also applies to products of more functions. Indeed, we can write #x^n# as a product of continuous functions on #\ivoo{0}{\infty}# : \[n\gt0:\qquad x^n=\underbrace{x\cdot x\cdots x}_{n\text{ terms}}\qquad n\lt0:\qquad x^n=\underbrace{\frac{1}{x}\cdot\frac{1}{x}\cdots\frac{1}{x}}_{n\text{ terms}}\tiny.\]

When #d# has the form #1/m# for an integer #m\in\mathbb{Z}\backslash\{0\}#, then we can write #x^d# as #\sqrt[m]{x}#. From this we know that she is continuous on #\ivoo{0}{\infty}#. If #d# is of the form #n/m#, we can apply rule 1 again. Then it is a product of root functions: \[\qquad x^{n/m}=\underbrace{\sqrt[m]{x}\cdot \sqrt[m]{x}\cdots \sqrt[m]{x}}_{n\text{ terms}}\qquad\tiny,\], hence, also continuous on #\ivoo{0}{\infty}#.