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Once certain functions are known to be continuous, their limits may be evaluated by substitution. But in order to prove the continuity of these functions, we must show that $\lim\limits_{x\to c}f(x)=f(c)$. To do this, we will need to construct delta-epsilon proofs based on the definition of the limit. Recall that the definition of the two-sided limit is:
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$\lim\limits_{x\to c} f(x)=L$ means that for every $\epsilon>0$, there exists a $\delta>0$, such that for every $x$, the expression $0< |x-c|<\delta$ implies $|f(x)-L|<\epsilon$. |
For each proof, we also provide a running commentary.
| Suppose $\epsilon>0$ has been provided. | A delta-epsilon proof requires an arbitrary epsilon. |
| Choose $\delta=1$. | We must show that there exists a delta for which the limit statement follows, and we claim this delta will suffice. |
| Then for all $x$, the expression $0< |x-c| <\delta$ | The beginning of the chain of implications |
| will imply $|k-k|=0$ | The statement is true under any set of conditions, so it really did not require the previous expression. But nevertheless, whenever the previous expression is true, this result is also true. |
| and thus $|k-k|<\epsilon$. | This is true since $\epsilon>0$. |
| Therefore, $\lim\limits_{x\to c}k=k$. | Having fulfilled the requirements of the definition of the limit, this statement results. |
| If $m=0$, the function becomes a constant function, whose limit was proved previously. Therefore, we assume $m\ne 0$. | The proof actually requires two cases, but the case where $m=0$ was previously proven. So we concentrate only on the second case. |
| Let $\epsilon>0$ be given. | A delta-epsilon proof requires an arbitrary epsilon. |
| Define $\delta=\dfrac{\epsilon}{|m|}$. Note that $\delta>0$. | A delta-epsilon proof must exhibit a delta that allows the chain of implications required by the definition to proceed. We claim this is the delta, and it is always positive. |
| Then for all $x$, the expression $0< |x-c| <\delta$ implies | Here is the beginning of the chain of implications. |
| $|x-c|<\dfrac{\epsilon}{|m|}$, $|mx-mc|<\epsilon$, and $|(mx+b)-(mc+b)|<\epsilon$ |
First, we replace $\delta$ by the value we gave it. Then we can multiply both sides by $|m|$. Since the product of absolute values is equal to the absolute value of the product, we can distribute. Then we add and subtract $b$ from the two terms inside the absolute values. |
| Therefore, $\lim\limits_{x\to c}(mx+b)=mc+b$. | The previous inequality was the necessary conclusion for the case $m\ne 0$. Both cases have now been proven, so we have demonstrated the truth of this limit statement. |
Assume that $r$ and $s$ are integers with no common factors (other than 1), and $s>1$. The following statements will be true.
The restrictions in the different cases are related to the domain of the function, and generally whenever the function is defined, it is continuous there. However, when the domain of the function is $[0,\infty)$, the power function will not exhibit two-sided continuity at zero (even though the function could be evaluated there).
| If $n=1$, this is a linear function, and is therefore continuous everywhere. | The continuity follows from the proof above that linear functions are continuous. |
| If $n>1$ is a positive integer, then we have $\lim\limits_{x\to c}x^n=\lim\limits_{x\to c}(x\cdots x)$. | We can rewrite the function as a product of $n$ factors. |
| Then the Product Law of limits gives $\lim\limits_{x\to c}x^n=\left(\lim\limits_{x\to c}x\right) \cdots \left(\lim\limits_{x\to c}x\right)=c\cdots c=c^n$ | This completes the proof for the first bullet (all positive integers). |
| If $n=0$, then the function $f(x)=x^n$ is equal to the constant function $f(x)=1$ at every real number except zero. Therefore, by the continuity of constant functions, this function is continuous everywhere except at zero. | We should note that the limit of this expression does exist as $x$ approaches zero, but since $0^0$ is undefined, the limit cannot be obtained by substitution. The All But One Point Theorem could be used to find its limit. |
| If $n>1$ is an even positive integer, then the function $f(x)=x^n$ is a strictly increasing function on the interval $[0,\infty)$. Therefore its inverse $f^{-1}(x)=x^{\frac{1}{n}}$ will produce $\lim\limits_{x\to c} x^{\frac{1}{n}}=c^{\frac{1}{n}}$ whenever $c>0$. | This is a result of the inverse law for limits, together with the continuity of the integer power function. This is sufficient to prove the continuity of $f(x)=\sqrt{x}$. |
| Similarly, if $n>1$ is an odd positive integer, then the function $f(x)=x^n$ is a strictly increasing function on the interval $(-\infty,\infty)$. Therefore its inverse $f^{-1}(x)=x^{\frac{1}{n}}$ will produce $\lim\limits_{x\to c} x^{\frac{1}{n}}=c^{\frac{1}{n}}$ for all real values of $c$. | This also is a result of the inverse law for limits, together with the continuity of the integer power function. |
| Now, if $n=\dfrac{r}{s}$, where $r$ and $s$ are positive integers with no common factors (other than 1), then we can write $f(x)=x^n=x^{\frac{r}{s}}=(x^r)^\frac{1}{s}$. By the Composition Limit Law, the continuity of this is established wherever the continuity of $f(x)=x^{\frac{1}{s}}$ existed. That is, $\lim\limits_{x\to c} x^{\frac{r}{s}}=c^{\frac{r}{s}}$ for all real values $c$ when $s$ is odd, and for values $c>0$ when $s$ is even. | This proves the third and fourth bullets for positive values of $n$. In short, the statement has now been established for all positive rational exponents. |
| If $n$ is a positive irrational number, we need to argue from the definition of the limit. Therefore, suppose $\epsilon>0$ has been given. | Now we look at the case with positive irrational exponents. As always, we begin our delta-epsilon proof with an arbitrary epsilon. |
| Define $\epsilon_1=\min\left\{\epsilon,\dfrac{c^n}{2}\right\}$ | To avoid difficulties that might occur if the original epsilon was too large, we choose a smaller epsilon where needed. Note that $\epsilon_1$ is positive as long as $c$ is positive. |
| and define $\delta=\min\left\{c-(c^n-\epsilon_1)^{\frac{1}{n}},(c^n+\epsilon_1)^{\frac{1}{n}}-c\right\}$. | Here is the delta that we claim will work. Notice that each delta candidate is positive. |
| Then for all $x$, whenever $0< |x-c| <\delta$, we will have | Here begins the chain of implications. |
| $-\delta < x-c < \delta$, $(c^n-\epsilon_1)^{\frac{1}{n}}-c < x-c < (c^n+\epsilon_1)^{\frac{1}{n}}-c$, $(c^n-\epsilon_1)^{\frac{1}{n}} < x < (c^n+\epsilon_1)^{\frac{1}{n}}$, $c^n-\epsilon_1 < x^n < c^n+\epsilon_1$, $-\epsilon_1 < x^n-c^n < \epsilon_1$, $|x^n-c^n|<\epsilon_1$, |
After replacing $\delta$ by the various quantities in its definition, to each expression we add $c$, raise to the power $n$, and subtract $c^n$. We can then write the inequality using absolute values. |
| and this implies $|x^n-c^n|<\epsilon$. | This results by the definition of $\epsilon_1$. |
| Therefore, $\lim\limits_{x\to c}x^n=c^n$ when $c$ is positive and $n$ is a positive irrational number. | This proves the sixth bullet for positive values of $n$, and establishes the result for all positive values of $n$. |
| Lastly, since $x^{-n}=\dfrac{1}{x^n}$, we can use the Quotient Limit Law to establish the theorem for all negative values of $n$. And therefore the entire theorem has been proven. | Thus, the second bullet is proved by the first bullet, the fifth bullet by the fourth bullet, and the negative portions of the third and sixth bullets by their respective positive results. |
| Suppose $P(x)=a_n x^n +a_{n-1}x^{n-1}+\ldots+a_1 x+a_0$. Then | Every polynomial can be written in this form. |
| $\lim\limits_{x\to c}P(x)=
\lim\limits_{x\to c}\left(a_n x^n +a_{n-1}x^{n-1}+\ldots+a_1 x+a_0\right)$ $=\lim\limits_{x\to c}a_n x^n+\lim\limits_{x\to c}a_{n-1}x^{n-1}+\ldots+\lim\limits_{x\to c}a_1 x +\lim\limits_{x\to c}a_0$ $=a_n\lim\limits_{x\to c}x^n+a_{n-1}\lim\limits_{x\to c}x^{n-1}+\ldots+a_1\lim\limits_{x\to c}x +a_0$ |
In the limit of the polynomial, we employed the Sum Limit Law, and the Scalar Multiple Limit Law. |
| $\lim\limits_{x\to c}P(x)=a_n c^n+a_{n-1}c^{n-1}+\ldots+a_1 c+a_0=P(c)$ | Then we can use the continuity of the Power Function (for positive integers) to establish the result for all real values of $c$. |
| Since $f(x)$ is a rational function, then $P(x)$ and $Q(x)$ are both polynomial functions. We obtain | This is the definition of a rational function. |
| $\lim\limits_{x\to c}f(x)=\lim\limits_{x\to c}\dfrac{P(x)}{Q(x)} =\dfrac{\lim\limits_{x\to c}P(x)}{\lim\limits_{x\to c}Q(x)}=\dfrac{P(c)}{Q(c)}=f(c)$ | Here we have used the Quotient Limit Law. That law requires that the denominator is not zero, but the restriction on $Q(c)$, together with the fact that $Q(x)$ is a polynomial, ensures that the restriction has been met. The continuity of polynomial functions does the rest. |
| Since $|x|=\sqrt{x^2}$, we can use the continuity of the functions $f(x)=x^2$ and $g(x)=\sqrt{x}$, together with the Composition Limit Law, to confirm the continuity of $|x|$ for every non-zero value of $c$. | Note that the continuity of the square root function did not extend to $x=0$, because the domain of the square root did not include any negative values. |
| To confirm that $\lim\limits_{x\to 0}|x|=0$, we note that $0\le |x|\le x^{\frac23}$ on the interval $[-1,1]$. | We have sandwiched the absolute value function between two functions whose continuity is already proven. |
| Then we have $\lim\limits_{x\to 0}0\le \lim\limits_{x\to 0}|x|\le \lim\limits_{x\to 0}x^{\frac23}$, which gives $0\le \lim\limits_{x\to 0}|x|\le 0$, and therefore $\lim\limits_{x\to 0}|x|=0$. | This is the result of the Sandwich Theorem. |
| Thus the absolute value function is continuous for all real numbers $c$. | Thus we have proven the theorem for all values of $c$. |