3
$\begingroup$

I just want to know, in calculating limits, when I do direct substitution, and it gives 3/0 instead of 0/0, does it mean for sure that the limit does not exist?

  • 0
    @Hurkyl: Agreed. What I usually try to do is to gauge the academic level the questioner is at, and try to answer in $a$precise way roughly at that level, perhaps pushing *a little* beyond.2012-09-16

1 Answers 1

7

Yes, it does mean that. Suppose $b_n\to 0$ and $c_n:=\frac{a_n}{b_n}\to L\in \mathbb R$. Then $\lim_{n\to\infty} a_n = \lim_{n\to\infty}b_n c_n=0\cdot L=0.$

  • 0
    @CameronBuie: Yes. Hagen is on the right way (+1). Thanks for lighting my mind.2012-09-15