# Limit as it approaches 0

1. Mar 7, 2015

### Tekneek

Why is the limit not just infinity?

wouldn't it be (1-infinity)/(1+infinity)?

2. Mar 7, 2015

### Staff: Mentor

Why should "-infinity/infinity" be infinity?

What is the limit of
$$\lim_{s \to 0} \frac{\frac{-1}{2}}{\frac{2}{s}}$$?

3. Mar 7, 2015

### Tekneek

-s/4, if s =0 then
Would't the limit be zero?

4. Mar 7, 2015

### Staff: Mentor

Sorry typo, this is the formula I meant:
$$\lim_{s \to 0} \frac{\frac{-1}{s}}{\frac{2}{s}}$$

But the more important part was the first question.

5. Mar 7, 2015

### Tekneek

For that formula it would be -1/2. I was thinking it would be infinity because you really can't put down a number with it unless maybe it is 1?

EDIT: Nvm go it. It ends up being -35/40. Thnx

Last edited: Mar 7, 2015
6. Mar 8, 2015

### Staff: Mentor

That is not the point. It follows the "-infinity/infinity" type. It is easy to simplify it here to see the limit is not infinity, so your original idea cannot work - that was the purpose of the example.
I don't understand that question.
Okay.

7. Mar 8, 2015

### PeroK

Whenever you write "infinity", you should write "a large number".

In this form (1 - a large number)/(1 + another large number) could be any negative number. It could be large and negative or it could be small and negative.

8. Mar 8, 2015

### HallsofIvy

A very important property of limits, not emphasized enough as I think it should be, is that "If f(x)= g(x) for all x except x= a, then $\lim_{x\to a} f(x)= \lim_{x\to a} g(x)$.

With this particular problem, we can, as long as n is not 0, multiply both numerator and denominator by n. That gives you $$\frac{s- 1}{s+ \frac{s+ 4}{7(s+ 5)}}$$. Now, take s= 0.

9. Mar 8, 2015

### Staff: Mentor

You lost two prefactors, HallsofIvy.

$$\frac{s- 1}{s+ \frac{10(2s+ 4)}{7(s+ 5)}}$$

10. Mar 8, 2015

### HallsofIvy

Thanks.

11. Mar 9, 2015

### Garrulo

The limit is of (s-1)/(s+(10*(2s+4)/7(s+5) that doing s=0 result in -7/15

12. Mar 9, 2015

### Staff: Mentor

No, check your algebra. The answer in post #1 (-7/8) is correct.