Continuity of a Function

DEFINITION OF CONTINUITY OF A FUNCTION AT A POINT. A function $f$ is said to be continuous a point $p$ if

(a) $f$ is defined at $p$, and

(b) MATH $f(x)=f(p).$

A function $f$ is continuous at $p$ if for every $\epsilon >0$ there is a $\delta >0$ such that
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EXAMPLE 1. Constant functions are continuous everywhere. If $f(x)=c$ for all $x$, then
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for every $p$, so $f$ is continuous everywhere.

EXAMPLE 2. The identity function is continuous everywhere. If $f(x)=x$ for all $x,$ we have
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for every $p$, so the identity function is continuous everywhere.

EXAMPLE 3. Let $f(x)=[x]$ for all $x.$ This function is continuous at every point $p$ which is not an integer. At the integers it is discontinuous, since the limit of $f$ does not exist, the right and left-hand limits being unequal. A discontinuity of this type, where the right- and left-hand limits exist but are unequal, is called a jump discontinuity.

EXAMPLE 4. The function $f$ for which $f(x)=1/x^2$ for $x\ne 0$,$f(0)=0$, is discontinuous at $0$. We say there is an infinite discontinuity at $0$ because the function takes arbitrarily large values near $0.$

EXAMPLE 5. Let $f(x)=1$ for $x\ne 0,f(0)=0$. This function is continuous everywhere except at $0$. It is discontinuous at $0$ because $f(0)$ is not equal to the limit of $f(x)$ as $x\rightarrow 0$. In this example, the discontinuity could be removed by redefining the function at $0$ to have the value $1$ instead of $0$. For this reason, a discontinuity of this type is called a removable discontinuity. Note that jump discontinuities, such as those possessed by the greatest-integer function, cannot be removed by simply changing the value of fat one point.

The Basic Limit Theorems

Calculations with limits may often be simplified by the use of the following theorem which provides basic rules for operating with limits.

Theorem. Let $f$ and $g$ be functions such that
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Then we have

(i) MATH

(ii) MATH

(iii)MATH

(iv)MATH if $B\ne 0.$

Proofs of the Basic Limit Theorem

Proofs of (i) and (ii). Since the two statements
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are equivalent, and since we have
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it suffices to prove part (i) of the theorem when the limits $A$ and $B$ are both zero. Suppose, then, that $f(x)\rightarrow 0$ and $g(x)\rightarrow 0$ as $x\rightarrow p$. We shall prove that MATH as $x\rightarrow p$. This means we must show that for every $\epsilon >0$ there is a $\delta >0$ such that
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Let $\epsilon >0$ be given. Since $f(x)\rightarrow 0$ as $x\rightarrow p$, there is a $\delta _1>0$ such that
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Similarly, $g(x)\rightarrow 0$ as $x\rightarrow p$, there is a $\delta _2>0$ such that
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If we let $\delta $ denote the smaller of the two numbers $\delta _1$ and $\delta _2$, then the above two inequalities are valid if MATH and hence, by the triangle inequality, we find that
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This proves (i). The proof of (ii) is entirely similar, except that the last step we use the inequality
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Proof of (iii). Suppose that we have proved part (iii) for the special case in which one the limits is $0$. Then the general case follows easily from this special case. In fact, all we need to do is write
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The special case implies that each term on the right approaches $0$ as $x\rightarrow p$ and, by property (i), the sum of the two terms also approaches $0$. Therefore, it remains to prove (iii) in the special case where one of the limits, say $B$, is $0$.

Suppose, then, that $f(x)\rightarrow A$ and $g(x)\rightarrow 0$ as $x\rightarrow p$. We wish to prove that MATH as $x\rightarrow p.$ To do this we must show that if a positive $\epsilon $ is given, there is a $\delta >0$ such that
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Since $f(x)\rightarrow A$ as $x\rightarrow p$, there is a $\delta _1>0$ such that
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For such $x$,. we have
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and hence
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Since $g(x)\rightarrow 0$ as $x\rightarrow p$, for every $\epsilon >0$ there is a $\delta _2$ such that
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Therefore, if we let $\delta $ be the smaller of the two numbers $\delta _1\,\,$and $\delta _2$, then both inequalities are valid whenever MATH, and for such $x$ we deduce the desired inequality
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This completes the proof of (iii).

Proof of (iv). Since the quotient $f(x)/g(x)$ is the product of $f(x)/B$ with $B/g(x)$, it suffices to prove that MATH as $x\rightarrow p$ and then appeal to (iii). Let $h(x)=g(x)/B$. Then $h(x)\rightarrow 1$ as $x\rightarrow p$, and we wish to prove that MATH as $x\rightarrow p$. Let $\epsilon >0$ be given. We must show that there is a $\delta >0$ such that
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The difference to be estimated may be written as follows.
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Since $h(x)\rightarrow 1$ as $x\rightarrow p$, we can choose a $\delta >0$ such that both inequalities
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are satisfied whenever MATH. The second of these inequalities implies $h(x)>1/2$ so MATH for such x. Using this in
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along with the inequality MATH, we obtain
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This completes the proof of (iv).

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