Calculus/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:
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 $0<\left|x-c\right|<\delta$.