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Given $\lim_{x\rightarrow a} f(x)=l$ and $\lim_{x\rightarrow a} g(x)=m$, I tried to prove that $lim_{x\rightarrow a} f(x)g(x)=lm$. Although the statement is simpler, proof is not obvious, and I faced some minor problems while writing the proof. Let $\epsilon >0$ be given. We have to find $\delta>0$ such that if $0<|x-a|<\delta$ then $|f(x)g(x)-lm|<\epsilon$. Now, $|f(x)g(x)-lm|\leq |g(x)|.|f(x)-l|+|l|.|g(x)-m|$, and it is sufficient to make the two terms on RHS less than $\epsilon/2$ for suitable $\delta$.

As limit of $g(x)$ is $m$, we can find $\delta_1>0$ such that

(*) $|g(x)-m|<\frac{\epsilon}{2(1+|l|)}$ whenever $0<|x-a|<\delta_1$.

Here we added $1$ to $l$ in the denominator since $l$ could be zero.

For $0<|x-a|<\delta_1$, we have $|g(x)|\leq |g(x)-m|+|m|<\frac{\epsilon}{2(1+|l|)}+|m|=C$ say. Here $C>0$.

Now, as limit of $f(x)$ is $l$, we can find $\delta_2>0$ such that $|f(x)-l|<\frac{\epsilon}{2C}$ whenever $0<|x-a|<\delta_2$.

Now one can easily show that if $\delta=min(\delta_1,\delta_2)$, then for $0<|x-a|<\delta$, both the terms $|g(x)|.|f(x)-l|$ and $|l|.|g(x)-m|$ are less than $\epsilon/2$, hence $|f(x)g(x)-lm|<\epsilon$.

My questions are as follows:

Is it necessary to consider case $l=0$ separately? Because After choosing $\delta_1$ such that (*) holds, I tried to show that $|l|.|g(x)-m|<\epsilon/2$ for $0<|x-a|<\delta_1$, which follows easily if $l\neq 0$. Does it follow easily if $l=0$?

In the proof, should we choose $\delta_2>0$ which is less than or equal to $\delta_1$ or we can choose $\delta_2$ independent of $\delta_1$?

Groups
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  • For your first question, the statement is trivially true when l=0. For your second question, you should remember that we do not choose deltas. We find deltas, given epsilons. There is no interaction between the deltas, because the limits of $f(x)$ and $g(x)$ exist independently. – Philip Hoskins Sep 04 '14 at 09:05

4 Answers4

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Note that the following identity can simplify the proof: $$f(x)g(x)-lm=(f(x)-l)(g(x)-m)+m(f(x)-l)+l(g(x)-m)$$ If you take $0<\varepsilon <1$ and $\delta>0$ as indicated by Michael Shi, you get immediately $$|f(x)g(x)-lm|\leq \varepsilon(\varepsilon+|m|+|l|)\leq \varepsilon(1+|m|+|l|)$$ and you are done without particular cases.

Kelenner
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If $|a-l|<\epsilon$ and $|b-m|<\epsilon$, then $$|ab-lm| = |a(b-m) + m(a-l)|<|a||b-m| + |m||a-l|$$ meaning that if $a$ is sufficiently close to $l$ and $b$ is close to $m$, then $ab$ will be close to $lm$.

5xum
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  • I understand the meaning vaguely, but I faced minor problems in "Writing the Proof" neatly. – Groups Sep 04 '14 at 08:23
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Regarding your second question, we are using the definition of $\lim_{x \rightarrow a} f(x) = l$ and $\lim_{x \rightarrow a} g(x) = m$ to find our $\delta_1$ and $\delta_2$. It's not so much about choosing our $\delta_2$ in relation to $\delta_1$ but rather using the information we are given. In choosing our $\delta = \min (\delta_1, \delta_2)$, we are necessarily forcing both conditions $$|x - a| < \delta \leq \delta_1 \implies |f(x) - l| < \epsilon$$ and $$|x - a| < \delta \leq \delta_2 \implies |g(x) - m| < \epsilon$$ to be true. I may be misinterpreting your question though.

Michael Shi
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  • I tried to explain that $\delta_2$ is independent of $\delta_1$, and proceeded in your way, but students were not satisfied. – Groups Sep 04 '14 at 08:25
  • I guess another way of explaining this is that we are unraveling what we know...we are given the $\delta_1$ and $\delta_2$ by the information we are given (the limits) and we can use that then to choose a proper $\delta$ which satisfies both inequalities. There's no such thing as choosing $\delta_1$ and $\delta_2$ in this case. – Michael Shi Sep 04 '14 at 08:29
  • $\delta_1$ and $\delta_2$ are the "values" or "bounds" we can use to know how small our $\delta$ has to be to work for a given $\epsilon$... – Michael Shi Sep 04 '14 at 08:31
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You can prove this by the following way which I feel is simpler :

$\lim_{x\to a}f(x)=l$, so $\forall \epsilon>0,\exists \delta_1>0$ such that $0<|x-a|<\delta_1 \implies|f(x)-l|<\epsilon$. Similarly

$\lim_{x\to a}g(x)=m$, so $\forall \epsilon>0,\exists \delta_2>0$ such that $0<|x-a|<\delta_2 \implies|g(x)-m|<\epsilon$. So,

letting $\delta=min(\delta_1,\delta_2)$ and 0<|x-a|< $\delta$,

$|f(x)g(x)-lm|=|f(x)g(x)-mf(x)+mf(x)-lm|=|f(x)(g(x)-m)+m(f(x)-l)|$ $$ \le|f(x)||g(x)-m|+m|f(x)-l|$$ $$ \lt\epsilon(|f(x)|+m)$$

which proves the result. $\square$

Sorry I could not format it properly.

creative
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