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I am a bit confused about application of algebra of limits in solving problems. I know that $\lim f(x)g(x) = \lim f(x) \cdot \lim g(x)$.

While solving problems on limits I partitioned the function in two such parts so that their individual limits can easily be found. But I didn't get right answer. (For example, I get a finite limit of first function, and limit of second part of function doesn't exist. But the limit of whole function exists finitely.)

So how to know if this algebra can be used to find a particular limit or not?

PrincessEev
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Mathaddict
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    You may use that $\lim_{n\to\infty} a_n\cdot b_n = \lim_{n\to\infty} a_n \cdot \lim_{n\to\infty} b_n$ whenever $\lim_{n\to\infty} a_n$ and $\lim_{n\to\infty} b_n$ exist. – bubububub Nov 26 '18 at 06:29

3 Answers3

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The limit of the product of functions equals the product of the limits of the functions only if both of those latter limits converge to some finite number. That is to say,

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

only if $\lim_{x \to c} f(x)$ and $\lim_{x \to c} g(x)$ are some finite number. (It is this existence of the limits part that is essential. There are ways to handle cases where $f,g$ approach infinity as well, but it's not through the above equality.)

(A minor note: a lot of similar laws, invoking for example the sum and difference of functions, also rely on this premise.)

PrincessEev
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  • Fair point, I think I more or less corrected that nuance. – PrincessEev Nov 26 '18 at 06:44
  • limit [(1 - sinx)/(pi - 2x)^4)(cosx)(8x^3 - pi^3)] as x tends to pi/2. Here the limit of (1-sinx)/(pi-2x)^4 is some non zero finite number. And the limit of rest part of function is 0. Then why the limit of the whole function is not zero? – Mathaddict Nov 26 '18 at 07:05
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We can use

$$\lim_{x\to x_0} f(x)\cdot g(x)=\lim_{x\to x_0} f(x)\cdot \lim_{x\to x_0} g(x)$$

when both limits exist finite otherwise we can also conclude directly when one is finite $L\neq0$ and the other infinite or when both are infinite using the rules

  • $L \cdot \infty= \infty$
  • $\infty \cdot \infty= \infty$

selecting the sign accordingly.

user
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The fact that you decomposed your function into a product with one undetermined factor does not question the truth of the statement all the other answers and comment point out. Just an unfortunate choice of writing your function. Let for instance the inverse function $$\frac 1 x$$

It converges, but if you write it as $$\frac{e^{ix}}{x} e^{-ix}$$

then the first factor converges to zero and the second one does not converge. The problem is not the algebras on limits (it applies whenever both limits do exist).

Desiderius Severus
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