Product Rule of Limits: Proof, Examples [Epsilon-Delta Method]

The product rule of limits says that the limit of the product of two functions is the same as the product of the limits of the individual functions. In this post, we will prove the product law of limits by the epsilon-delta method.

Question: What is an epsilon-delta proof of the product law of limits? Here, you will find the answer given below.

Let us now recall the epsilon-delta definition of limit: Let $f(x)$ be a function and $\lim\limits_{x \to a} f(x) = A$. Then for every $\epsilon>0$, there exists a $\delta>0$ such that whenever $0<|x-a|<\delta$ we have that

$|f(x)-A| < \epsilon$ …(*)

Product Law of Limits

If the limits of two functions f and g exist when x tends \to a, then

$\lim\limits_{x \to a}(fg)$ $=\lim\limits_{x \to a} f \cdot  \lim\limits_{x \to a} g$

Proof:

Step 1: To prove the product law of limits, we will use the epsilon-delta method. Let $\lim\limits_{x \to a} f(x) = A$ and $\lim\limits_{x \to a} g(x) = B$.

Step 2: Now,

$|f(x)g(x) – AB|$ $= |f(x)g(x) – Bf(x) + Bf(x) – AB|$

$= |f(x) (g(x) – B) + B (f(x) – A)|$

$\leq |f(x) (g(x) – B)+ |B(f(x) – A)|$ by the triangle inequality |a+b| ≤ |a|+|b|

$= |f(x)|  |g(x)- B| + B|f(x) – A|$

Step 3: Thus, we have shown that

$|f(x)g(x) – AB|$  $\leq  |f(x)|  |g(x)- B| + B|f(x) – A|$ …(**)

Now, by (*) we have that for $\epsilon=1$, there is a $\delta>0$ such that

$|f(x)-A|<1$ whenever $0<|x-a|<\delta_1$ …(I)

Hence, $|f(x)| = |f(x)-A+A|$ $\leq |f(x)-A| + |A|$ $< 1+|A|$ …(II)

Step 4: As $\lim\limits_{x \to a} f(x)=A$,  for $\epsilon=\dfrac{\epsilon}{2(1+|B|)}>0$, there is a $\delta_2>0$ such that

$|f(x)-A|<\dfrac{\epsilon}{2(1+|B|)}$ whenever $0<|x-a|<\delta_2$ …(III)

On the other hand, as $\lim\limits_{x \to a} g(x)=B$,  for $\epsilon=\dfrac{\epsilon}{2(1+|A|)}>0$, there is a $\delta_3>0$ such that

$|g(x)-B|<\dfrac{\epsilon}{2(1+|A|)}$ whenever $0<|x-a|<\delta_3$…(IV)

Step 5: Set $\delta:={\delta_1, \delta_2, \delta_3}$. If  $0<|x-a|<\delta$ then we have from (**) that

$|f(x)g(x) – AB|$ $\leq |f(x)|  |g(x)- B| + B|f(x) – A|$

$< |f(x)|  |g(x)- B| + (1+|B|) |f(x) – A|$ $< (1+|B|) \cdot \dfrac{\epsilon}{2(1+|A|)}$  $+(1+|B|) \cdot \dfrac{\epsilon}{2(1+|B|)}$, by (II), (III) and (IV)

$< \dfrac{\epsilon}{2} +\dfrac{\epsilon}{2}$

$= \epsilon$

Conclusion: Therefore, by the epsilon-delta definition, we have shown that $\lim\limits_{x \to a} [f(x)\cdot g(x)]$ $= AB = \lim\limits_{x \to a} f(x) \cdot \lim\limits_{x \to a}g(x)$. This proves the product rule of limits.

Also Read:

Epsilon delta Definition of limit | Negation of epsilon-delta Definition

Sum rule of limits: Proof and Examples

Examples

We know that $\lim\limits_{x\to 0} x^2=0$ and $\lim\limits_{x\to 0} (x+1)=1$.

So by the above product rule of limits, the limit of $x^2(x+1)$ when x tends to 0 is equal to

$\lim\limits_{x\to 0} x^2(x+1)$

= $\lim\limits_{x\to 0} x^2$ $\cdot \lim\limits_{x\to 0} (x+1)$

= 0 × 1

= 0

Also Read:

Integration of Cube Root of x

Derivative of root(1+x)

Derivative of log(cos x)

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