Last modified on 15 July 2012, at 21:26

Calculus/Proofs of Some Basic Limit Rules

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Proofs of Some Basic Limit Rules

Now that we have the formal definition of a limit, we can set about proving some of the properties we stated earlier in this chapter about limits.

Constant Rule for Limits

If b and c are constants then  \lim_{x\to c} b = b.

Proof of the Constant Rule for Limits:
To prove that  \lim_{x\to c} f(x) = b, we need to find a \delta>0 such that for every \varepsilon>0, \left|b-b\right|<\varepsilon whenever \left|x-c\right|<\delta. \left|b-b\right|=0 and \varepsilon>0, so \left|b-b\right|<\varepsilon is satisfied independent of any value of \delta; that is, we can choose any \delta we like and the \varepsilon condition holds.

Identity Rule for Limits

If c is a constant then  \lim_{x\to c} x = c.

Proof of the Identity Rule for Limits:
To prove that  \lim_{x\to c} x = c, we need to find a \delta>0 such that for every \varepsilon>0, \left|x-c\right|<\varepsilon whenever \left|x-c\right|<\delta. Choosing \delta=\varepsilon satisfies this condition.

Scalar Product Rule for Limits

Suppose that \lim_{x\to c} f(x) =L for finite L and that k is constant. Then  \lim_{x\to c} k f(x) = k \cdot \lim_{x\to c} f(x) =  k L

Proof of the Scalar Product Rule for Limits:
Since we are given that \lim_{x\to c} f(x) =L, there must be some function, call it \delta_{f}(\varepsilon), such that for every \varepsilon>0, \left|f(x)-L\right|<\varepsilon whenever \left|x-c\right|<\delta_{f}(\varepsilon). Now we need to find a \delta_{kf}(\varepsilon) such that for all \varepsilon>0, \left|k f(x)-k L\right|<\varepsilon whenever \left|x-c\right|<\delta_{kf}(\varepsilon).
First let's suppose that k>0. \left|k f(x)-k L\right| = k \left|f(x)-L\right|<\varepsilon, so \left|f(x)-L\right|<\varepsilon/k. In this case, letting \delta_{kf}(\varepsilon)=\delta_{f}(\varepsilon/k) satisfies the limit condition.
Now suppose that k=0. Since f(x) has a limit at x=c, we know from the definition of a limit that f(x) is defined in an open interval D that contains c (except maybe at c itself). In particular, we know that f(x) doesn't blow up to infinity within D (except maybe at c, but that won't affect the limit), so that  0 f(x)=0 in D. Since k f(x) is the constant function 0 in D, the limit  \lim_{x\to c} k f(x) = 0 by the Constant Rule for Limits.
Finally, suppose that k<0. \left|k f(x)-k L\right| = -k \left|f(x)-L\right|<\varepsilon, so \left|f(x)-L\right|<-\varepsilon/k. In this case, letting \delta_{kf}(\varepsilon)=\delta_{f}(-\varepsilon/k) satisfies the limit condition.

Sum Rule for Limits
Suppose that \lim_{x\to c} f(x) =L and \lim_{x\to c} g(x) =M. Then

 \lim_{x\to c} [f(x) + g(x)] = \lim_{x\to c} f(x) +  \lim_{x\to c} g(x) =  L + M

Proof of the Sum Rule for Limits:
Since we are given that \lim_{x\to c} f(x) =L and \lim_{x\to c} g(x) =M, there must be functions, call them \delta_{f}(\varepsilon) and \delta_{g}(\varepsilon), such that for all \varepsilon>0, \left|f(x)-L\right|<\varepsilon whenever \left|x-c\right|<\delta_{f}(\varepsilon), and .\left|g(x)-M\right|<\varepsilon whenever \left|x-c\right|<\delta_{g}(\varepsilon).
Adding the two inequalities gives \left|f(x)-L\right| + \left|g(x)-M\right| < 2\varepsilon. By the triangle inequality we have \left|f(x)-L\right| + \left|g(x)-M\right| \geq \left|(f(x)-L)+(g(x)-M)\right|=\left|(f(x)+g(x))-(L+M)\right|, so we have \left|(f(x)+g(x))-(L+M)\right|<2\varepsilon whenever \left|x-c\right|<\delta_{f}(\varepsilon) and \left|x-c\right|<\delta_{g}(\varepsilon). Let \delta_{fg}(\varepsilon) be the smaller of \delta_{f}(\varepsilon/2) and \delta_{g}(\varepsilon/2). Then this \delta satisfies the definition of a limit for \lim_{x\to c} [f(x) + g(x)] having limit  L + M .

Difference Rule for Limits
Suppose that \lim_{x\to c} f(x) =L and \lim_{x\to c} g(x) =M. Then

 \lim_{x\to c} [f(x) - g(x)] = \lim_{x\to c} f(x) -  \lim_{x\to c} g(x) =  L - M

Proof of the Difference Rule for Limits: Define h(x)=-g(x). By the Scalar Product Rule for Limits, \lim_{x\to c}h(x)=-M. Then by the Sum Rule for Limits, \lim_{x\to c}(f(x)-g(x))=\lim_{x\to c}(f(x)+h(x))=L-M.

Product Rule for Limits
Suppose that \lim_{x\to c} f(x) =L and \lim_{x\to c} g(x) =M. Then

 \lim_{x\to c} [f(x) g(x)] = \lim_{x\to c} f(x) \lim_{x\to c} g(x) = L M

Proof of the Product Rule for Limits:[1]
Let \varepsilon be any positive number. The assumptions imply the existence of the positive numbers \delta_{1}, \delta_{2}, \delta_{3} such that

(1)\qquad\left|f(x)-L\right|<\frac{\varepsilon}{2(1+\left|M\right|)} when 0<\left|x-c\right|<\delta_{1}
(2)\qquad\left|g(x)-M\right|<\frac{\varepsilon}{2(1+\left|L\right|)} when 0<\left|x-c\right|<\delta_{2}
(3)\qquad\left|g(x)-M\right|<1 when 0<\left|x-c\right|<\delta_{3}

According to the condition (3) we see that

\left|g(x)\right|=\left|g(x)-M+M\right|\leq\left|g(x)-M\right|+\left|M\right|<1+\left|M\right| when 0<\left|x-c\right|<\delta_{3}

Supposing then that 0<\left|x-c\right|<\min\{\delta_{1},\delta_{2},\delta_{3}\} and using (1) and (2) we obtain

\begin{align}\left|f(x)g(x)-LM\right|&=\left|f(x)g(x)-Lg(x)+Lg(x)-LM\right|\\
&\leq\left|f(x)g(x)-Lg(x)\right|+\left|Lg(x)-LM\right|\\
&=\left|g(x)\right|\cdot\left|f(x)-L\right|+\left|L\right|\cdot\left|g(x)-M\right|\\
&<(1+\left|M\right|)\frac{\varepsilon}{2(1+\left|M\right|)}+(1+\left|L\right|)\frac{\varepsilon}{2(1+\left|L\right|)}\\
&=\varepsilon
\end{align}

Quotient Rule for Limits
Suppose that \lim_{x\to c} f(x) =L and \lim_{x\to c} g(x) =M and M \neq 0. Then

 \lim_{x\to c} \frac{f(x)}{g(x)} = \frac{\lim_{x\to c} f(x)}{\lim_{x\to c} g(x)} = \frac{L}{M}

Proof of the Quotient Rule for Limits:
If we can show that \lim_{x\to c}\frac{1}{g(x)}=\frac{1}{M}, then we can define a function, h(x) as h(x)=\frac{1}{g(x)} and appeal to the Product Rule for Limits to prove the theorem. So we just need to prove that \lim_{x\to c}\frac{1}{g(x)}=\frac{1}{M}.
Let \varepsilon be any positive number. The assumptions imply the existence of the positive numbers \delta_{1}, \delta_{2} such that

(1)\qquad\left|g(x)-M\right|<\varepsilon\left|M\right|(1+\left|M\right|) when 0<\left|x-c\right|<\delta_{1}
(2)\qquad\left|g(x)-M\right|<1 when 0<\left|x-c\right|<\delta_{2}

According to the condition (2) we see that

\left|g(x)\right|=\left|g(x)-M+M\right|\leq\left|g(x)-M\right|+\left|M\right|<1+\left|M\right| when 0<\left|x-c\right|<\delta_{2}

which implies that

(3)\qquad\left|\frac{1}{g(x)}\right|>\frac{1}{1+\left|M\right|} when 0<\left|x-c\right|<\delta_{2}

Supposing then that 0<\left|x-c\right|<\min\{\delta_{1},\delta_{2}\} and using (1) and (3) we obtain

\begin{align}\left|\frac{1}{g(x)}-\frac{1}{M}\right|&=\left|\frac{M-g(x)}{Mg(x)}\right|\\
&=\left|\frac{g(x)-M}{Mg(x)}\right|\\
&=\left|\frac{1}{g(x)}\right|\cdot\left|\frac{g(x)-M}{M}\right|\\
&<\frac{1}{1+\left|M\right|}\cdot\left|\frac{g(x)-M}{M}\right|\\
&<\frac{1}{1+\left|M\right|}\cdot\left|\frac{\varepsilon\left|M\right|(1+\left|M\right|)}{M}\right|\\
&=\varepsilon
\end{align}
Theorem: (Squeeze Theorem)
Suppose that g(x) \le f(x) \le h(x) holds for all x in some open interval containing c, except possibly at x=c itself. Suppose also that \lim_{x\to c}g(x)=\lim_{x\to c}h(x)=L. Then \lim_{x\to c}f(x)=L also.

Proof of the Squeeze Theorem:
From the assumptions, we know that there exists a \delta such that \left|g(x)-L\right|<\varepsilon and \left|h(x)-L\right|<\varepsilon when 0<\left|x-c\right|<\delta.
These inequalities are equivalent to L-\varepsilon<g(x)<L+\varepsilon and L-\varepsilon<h(x)<L+\varepsilon when 0<\left|x-c\right|<\delta.
Using what we know about the relative ordering of f(x), g(x), and h(x), we have
L-\varepsilon<g(x)<f(x)<h(x)<L+\varepsilon when 0<\left|x-c\right|<\delta.
or
-\varepsilon<g(x)-L<f(x)-L<h(x)-L<\varepsilon when 0<\left|x-c\right|<\delta.
So
\left|f(x)-L\right|<max(\left|g(x)-L\right|,\left|h(x)-L\right|)<\varepsilon when 0<\left|x-c\right|<\delta.

NotesEdit

  1. This proof is adapted from one found at planetmath.org/encyclopedia/ProofOfLimitRuleOfProduct.html due to Planet Math user pahio and made available under the terms of the Creative Commons By/Share-Alike License.
← Formal Definition of the Limit Calculus Limits/Exercises →
Proofs of Some Basic Limit Rules