Proof of a Limit Law: If lim f(x) = L, lim g(x) = M, then...

[o_O]

Question: If \(\displaystyle \lim_{x \to a} f(x) = L\) and \(\displaystyle \lim_{x \to a} g(x) = M\), then \(\displaystyle \lim_{x \to a} (f(x)g(x)) = LM\).

Proof from James Stewarts':

Let \(\displaystyle \epsilon > 0\) . We want to find \(\displaystyle \delta > 0\) such that \(\displaystyle |f(x)g(x) - LM|\) whenever \(\displaystyle 0 < |x - a| < \delta\).

\(\displaystyle \left| f(x)g(x) - LM \right|\)

\(\displaystyle = \left| f(x)g(x) - Lg(x) + Lg(x) - LM \right|\)

\(\displaystyle = \left|\left[f(x) - L\right]g(x) + L\left[g(x) - M\right]\right|\)

\(\displaystyle \leq \left|\left[f(x) - L\right]\left(g(x)\right)\right| + \left|\left(L\right)\left[g(x) - M\right]\right|\) (Triangle inequality)

\(\displaystyle = \left|f(x) - L\right|\left|g(x)\right| + \left|L\right|\left|g(x) - M\right|\)

We want to make each of these terms less than \(\displaystyle \frac{\epsilon}{2}\).

Since \(\displaystyle \lim_{x \to a} g(x) = M\), there is a number \(\displaystyle \delta_{1} > 0\) such that: \(\displaystyle |g(x) - M| < \frac{\epsilon}{2\left(1 + |L|\right)}\)* whenever \(\displaystyle 0 < |x - a| < \delta_{1}\).

Also, there is \(\displaystyle \delta_{2}>0\) such that if \(\displaystyle 0 < |x - a| < \delta_{2}\), then \(\displaystyle |g(x) - M| < 1\) and therefore:

\(\displaystyle \left|g(x)\right| = \left|g(x) - M + M\right| \hspace{4mm} \leq \hspace{4mm} \left|g(x) - M\right| + \left|M\right| \hspace{4mm}< \hspace{4mm} 1 + \left|M\right|\)

Since \(\displaystyle \lim_{x \to a}f(x) = L\), there is a number \(\displaystyle \delta_{3}>0\) such that: \(\displaystyle \left|f(x) - L\right|<\frac{\epsilon}{2\left(1+|M|\right)}\)* whenever \(\displaystyle 0 < |x - a| < \delta_{3}\).

Let \(\displaystyle \delta = min\left\{\delta_{1},\delta_{2},\delta_{3}\right\}\). If \(\displaystyle 0 < |x - a| < \delta\).

Then we have \(\displaystyle 0 < |x-a| <\delta_{1}, \hspace{4mm} 0 < | x-a| < \delta_{2}, \hspace{4mm} 0 <|x-a|<\delta_{3}\)

so we can combine the inequalities to obtain:

\(\displaystyle \left|f(x)g(x) - LM\right|\)

\(\displaystyle \leq \hspace{4mm}\left|f(x)-L\right| \left|g(x)\right| \hspace{4mm}+ \hspace{4mm}\left|L\right|\left|g(x)-M\right|\)

\(\displaystyle = \frac{\epsilon}{2\left(1 + |M|\right)} \left(1 + |M|\right) \hspace{4mm} + \hspace{4mm} |L| \frac{\epsilon}{2\left(1 + |L|\right)\)

\(\displaystyle < \frac{\epsilon}{2} + \frac{\epsilon}{2} = \epsilon\)

Therefore, \(\displaystyle \lim_{x \to a} f(x)g(x) = LM\)

Problem: I don't exactly how they made the statements with the *s. Where did these inequalities come from and how can they be asserted? Also, I'm kind of iffy on why we need three \(\displaystyle \delta\)s ... Any help would be appreciated!

 

[pka]

Given the extent and fundamental nature of these questions, I think that you should sit down with a live tutor. These are important concepts to grasp.