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The limit of the product of two functions should be equal to the product of the limits:

$$\lim_{x\to\infty}f(x)g(x) = \lim_{x\to\infty}f(x) \lim_{x\to\infty}g(x)$$

Now, the limit of $\frac{(x-1)3}{4x}$ = $\frac{3}{4}$

But the limit of $\frac{x-1}{4}$ = $\infty$ and the limit of $\frac{3}{x}$ = 0

How is this? Thanks!

user3435407
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2 Answers2

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Well, you proved that result was false.

But when is it correct? First assume both $ \lim_{x \rightarrow \infty} f(x) $ and $ \lim_{x \rightarrow \infty} g(x) $ exist (on the extended real number line).

If both limits are finite, if both are infinite or if one is infinite and the other is non-zero, your theorem is correct and indeed

$$ \lim_{x \rightarrow \infty} f(x)g(x) = \lim_{x \rightarrow \infty} f(x) \lim_{x \rightarrow \infty} g(x) . $$

To verify this, simply use the definition of convergence and check all cases, using usual conventions to multiply infinite numbers. In total there are six cases to verify (finite/finite, $+\infty/+\infty$, $+\infty/-\infty$, $-\infty/-\infty$, non-zero/$+\infty$, non-zero/$-\infty$).

Use the following definition for convergence: $\lim_{x \rightarrow +\infty} f(x) = l$ if for every neighborhood $W$ of $l$, there exists a neighborhood $V$ of $+\infty$ such that for every $x$ in $V$, $f(x)$ is in $W$.

In the last definition, the neighborhood of a number is simply a non-empty open set containing that number.

Axel
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The two expressions are equal when both of the limits on the right side exist and are finite numbers.

Likewise $\displaystyle \lim_{x\to a}\frac{f(x)}{g(x)} = \frac{\lim_{x\to a} f(x)}{\lim_{x\to b} g(x)}$ if the limits in the numerator and denominator exist and are finite numbers and the limit in the denominator is not $0$. But it is crucially important to consider limits of the form on the left side in the case where the two limits on the right side are both $0$, because that is what derivatives are.

Michael Hardy
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    They are also equal if both limits exist, one is infinite, and the other is not zero. In other words, the only indeterminate form among products is $\infty \cdot 0$. – Ian Jul 15 '15 at 15:38
  • ah ok, I see, thank you! I just wanted to ask this possibility, cause in another case I saw a simplification for e*inf and that should work, giving inf. – user3435407 Jul 15 '15 at 15:42
  • What is $b$ in your equation? A mistake? If so, for the case where the limits in the RHS are both zero or both infinite, there's something useful called L'Hôpital's rule. – Stewart Jan 02 '23 at 15:16