Below we provide formal proofs of the statements for the case that #a#, #b# and #c# are real. The cases where one or two of these are #\pm\infty# can be proven in almost the same way. A small positive number #\epsilon# is always chosen. We then use the given limits to make suitable estimates (not always #\epsilon# itself) for the limit to be proven. In doing so, we assign a positive number #\delta# so that the estimates work for all #x# with #|x-a|\lt\delta#.

In the estimates, we use the inequalities #|x+y|\leq|x|+|y|#, #|x|-|y|\leq|x-y|# and the equality #x^n-y^n = (x-y)(x^{n-1}+x^{n-2}y+\cdots+xy^{n-2}+y^{n-1})#.

Rule 1: #\lim_{x\to a}\left(f(x)+g(x)\right)=b+c#.
Choose an arbitrary #\epsilon\gt0#. According to the definition of #\displaystyle\lim_{x\to a}f(x)=b#, there is a positive number #\delta_1# such that for all #x# with #0\lt|x-a|\lt\delta_1# it holds that #|f(x)-b|\lt\dfrac{\epsilon}{2}#. Similarly, from the definition of #\lim_{x\to a} g(x)=c#, there is a positive number #\delta_2# such that for all #x# with #0\lt|x-a|\lt\delta_2# it holds that #|g(x)-c|\lt\dfrac{\epsilon}{2}#. Let #\delta=\min(\delta_1,\delta_2)#, the minimum of #\delta_1# and #\delta_2#. Then #\delta\gt0# and for all #x# with #|x-a|\lt\delta# it holds:

\[\begin{array}{rcl} \left|\left(f(x)+g(x)\right)-(b+c)\right| & = & |\left(f(x)-b\right)+\left(g(x)-c\right)|\\ & \le & |f(x)-b|+|g(x)-c| \\ & < & \dfrac{\epsilon}{2}+\dfrac{\epsilon}{2}= \epsilon{\tiny.}\end{array}\]Thus, the limit of #f+g# at #a# is equal to #b+c#.

Rule 2: The proof of #\lim_{x\to a}\left(f(x)-g(x)\right)=b-c# follows in the same way as that of Rule 1. Note that #|g(x)-c|=|c-g(x)|#.

Rule 3: #\lim_{x\to a}\left(f(x)\cdot g(x)\right)=b\cdot c#.
Choose an arbitrary #\epsilon\gt0#. According to the definition of #\lim_{x\to a}f(x)=b#, there is a positive number #\delta_1# such that for all #x# with #0\lt|x-a|\lt\delta_1# it holds that #|f(x)-b|\lt\min\left(\dfrac{\epsilon}{2|c|+1},\dfrac{|b|+1}{2}\right)#. Similarly, from the definition of #\displaystyle\lim_{x\to a} g(x)=c#, there is a positive number #\delta_2# such that for all #x# with #0\lt|x-a|\lt\delta_2# it holds that #|g(x)-c|\lt\dfrac{\epsilon}{3|b|+1}#. Now take #\delta=\min(\delta_1,\delta_2)# and assume that #|x-a|\lt\delta#. Then the above estimates hold for #|f(x)-b|# and #|g(x)-c|#. Since from #|f(x)-b|\lt \dfrac{|b|+1}{2}# it follows that #|f(x)|\lt\dfrac{3|b|+1}{2}#, we find \[\begin{array}{rcl}\left|\left(f(x)\cdot g(x)\right)-(b\cdot c)\right|& = & |f(x) \left(g(x)-c\right)+\left(f(x)-b\right)c|\\& \le & |f(x)|\cdot |g(x)-c|+|f(x)-b|\cdot |c|\\ & < & \dfrac{3|b|+1}{2}\dfrac{\epsilon}{3|b|+1}+\dfrac{\epsilon}{2|c|+1}|c|\\ & \le & \dfrac{\epsilon}{2} +\dfrac{\epsilon}{2} =\epsilon \tiny.\end{array}\]Thus, the limit of #f\cdot g# at #a# is equal to #bc#.

Rule 4: Let #c\ne0#. We need to show that #\lim_{x\to a}\dfrac{f(x)}{g(x)}=\frac{b}{c}#.
We first prove #\displaystyle\lim_{x\to a}\dfrac{1}{g(x)}=\frac{1}{c}#. Choose #\epsilon\gt0# arbitrarily. We choose a positive number #\delta# such that for all #x# with #|x-a|\lt\delta# it holds that #|g(x)-c|\le\min(\frac{\epsilon |c|^2}{2},\dfrac{|c|}{2})#. From #|g(x)-c|\lt \dfrac{|c|}{2}# it follows that #|g(x)|\gt\dfrac{|c|}{2}#, so for #|x-a|\lt \delta#:\[\left|\frac{1}{g(x)}-\frac{1}{c}\right| = \frac{|g(x)-c|}{|g(x)|\cdot|c|} < \frac{|c|^2\frac{\epsilon}{2}}{\frac{|c|}{2}|c|} = \epsilon \tiny.\]

Thus, we have proven #\displaystyle\lim_{x\to a}\dfrac{1}{g(x)}=\frac{1}{c}#. The general rule follows from the previous one with #\dfrac{1}{g(x)}# in stead of #g(x)#.

Rule 5: If #d\ne0# is a rational number and #b\gt0#, then #\displaystyle\lim_{x\to a}f(x)^d=b^d#.

We have not formally defined the number #b^d# for arbitrary #d#, only for #d# a rational number. Therefore, we provide the rule and proof here only for rational numbers.

If #d\lt0#, then #\lim_{x\to a}f(x)^d=\lim_{x\to a}\dfrac{1}{f(x)^{-d}}#. If we have established that the rule holds for #d\gt0#, then we can further reduce this thanks to rule 3 to #\dfrac{1}{b^{-d}}=b^d#. Therefore, we can limit ourselves to a proof for the case #d\gt0#.

Since #d# is a rational number, we can write #d=\dfrac{n}{m}# for two natural numbers #n# and #m# with #m\gt0#. First, consider the case #m=1# and #n\geq1#. Since\[f(x)^n=\underbrace{f(x)\cdot f(x)\cdots f(x)}_{n\text{ terms}}\tiny,\]this case follows from repeated application of rule 3.

Now we consider the special case where #n=1#. Choose #\epsilon\gt0# arbitrarily. We write \[D=\left(\dfrac{b}{2}\right)^{\frac{m-1}{m}}+\left(\dfrac{b}{2}\right)^{\frac{m-2}{m}}b^{\frac{1}{m}}+\cdots +\left(\dfrac{b}{2}\right)^{\frac{1}{m}}b^{\frac{m-2}{m}}+b^{\frac{m-1}{m}}\tiny.\]This is a positive number. Using #\lim_{x\to a}f(x) = b#, we know that there is a positive number #\delta# such that for all #x# with #|x-a|\lt \delta#, it holds that #|f(x)-b|\lt \min\left(\epsilon D, \dfrac{b}{2}\right)#. For these #x#, we find #|f(x)|\ge \dfrac{b}{2}# and thus\[f(x)^{\frac{m-1}{m}}+f(x)^{\frac{m-2}{m}}b^{\frac{1}{m}}+\cdots +f(x)^{\frac{1}{m}}b^{\frac{m-2}{m}}+b^{\frac{m-1}{m}}\ge D\tiny.\]The left-hand side is equal to #\dfrac{f(x)-b}{f(x)^{\frac{1}{m}}-b^{\frac{1}{m}}}#. From this it follows\[\left|f(x)^{\frac{1}{m}}-b^{\frac{1}{m}}\right| = \dfrac{|f(x)-b|}{f(x)^{\frac{m-1}{m}}+f(x)^{\frac{m-2}{m}}b^{\frac{1}{m}}+\cdots +f(x)^{\frac{1}{m}}b^{\frac{m-2}{m}}+b^{\frac{m-1}{m}}}\le\dfrac{\epsilon D}{D}=\epsilon\tiny.\]

This proves that #\lim_{x\to a}f(x)^d=b^d# for #d=\dfrac{1}{m}#. By applying the case #d=n# to this result, we obtain the general case: #\lim_{x\to a}f(x)^d =\lim_{x\to a}\left(f(x)^{\frac{1}{m}}\right)^n = \left(\lim_{x\to a}f(x)^{\frac{1}{m}}\right)^n = \left(b^{\frac{1}{m}}\right)^n = b^\frac{n}{m} =b^d#.

Rule 6: If #h# is a function with #\lim_{x\to c}h(x) = a#, then #\displaystyle\lim_{x\to c}f(h(x)) = b#.
Choose #\epsilon\gt0# arbitrarily. Due to #\lim_{x\to a}f(x) = b#, there is a #\delta_f\gt0# such that for all #y# with #|y-a|\le\delta_f#, it holds that #|f(y)-b|\lt \epsilon#. Due to #\lim_{x\to c}h(x) = a#, there is a #\delta\gt0# such that for all #x# with #|x-c|\le\delta#, it holds that #|h(x)-a|\lt \delta_f#. Consequently, for all #x# with #|x-c|\le\delta#, it holds that #|h(x)-a|\lt\delta_f#, so #|f(h(x))-b|\lt\epsilon# (apply the previous with #y=h(x)#). This proves that #\lim_{x\to c}f(h(x)) = b#.