Proof of Various Limit Properties — Question 1

Question 1 - Solution

Let $\varepsilon>0$ be given.

We must show there exists $\delta>0$ such that if $0<|x-a|<\delta$, then

$$|f(x)g(x)-LM|<\varepsilon.$$

Start by adding and subtracting $L g(x)$:

$$f(x)g(x)-LM = f(x)g(x)-Lg(x)+Lg(x)-LM = g(x)\bigl(f(x)-L\bigr)+L\bigl(g(x)-M\bigr).$$

Take absolute values and use the triangle inequality:

$$|f(x)g(x)-LM| \le |g(x)|\,|f(x)-L|+|L|\,|g(x)-M|.$$

To control $|g(x)|$, use the fact that $g(x)\to M$.

Since $\lim_{x\to a}g(x)=M$, there exists $\delta_1>0$ such that if $0<|x-a|<\delta_1$, then

$$|g(x)-M|<1.$$

Then, for $0<|x-a|<\delta_1$,

$$|g(x)| = |g(x)-M+M| \le |g(x)-M|+|M| < 1+|M|.$$

Now choose $\delta_2>0$ such that if $0<|x-a|<\delta_2$, then

$$|f(x)-L|<\frac{\varepsilon}{2(1+|M|)}.$$

This is possible because $\lim_{x\to a}f(x)=L$.

Next choose $\delta_3>0$ such that if $0<|x-a|<\delta_3$, then

$$|g(x)-M|<\frac{\varepsilon}{2(1+|L|)}.$$

This is possible because $\lim_{x\to a}g(x)=M$.

Let

$$\delta=\min(\delta_1,\delta_2,\delta_3).$$

Then whenever $0<|x-a|<\delta$, we have simultaneously:

$$|g(x)|<1+|M|, \quad |f(x)-L|<\frac{\varepsilon}{2(1+|M|)}, \quad |g(x)-M|<\frac{\varepsilon}{2(1+|L|)}.$$

Substitute into the estimate:

$$|f(x)g(x)-LM| \le |g(x)|\,|f(x)-L|+|L|\,|g(x)-M|.$$

For the first term:

$$|g(x)|\,|f(x)-L| < (1+|M|)\cdot\frac{\varepsilon}{2(1+|M|)} =\frac{\varepsilon}{2}.$$

For the second term:

$$|L|\,|g(x)-M| \le \frac{\varepsilon|L|}{2(1+|L|)} <\frac{\varepsilon}{2}.$$

Therefore,

$$|f(x)g(x)-LM|<\frac{\varepsilon}{2}+\frac{\varepsilon}{2}=\varepsilon.$$

Since $\varepsilon>0$ was arbitrary, it follows that

$$\lim_{x\to a}\bigl(f(x)g(x)\bigr)=LM.$$