# Limit of (1-1)/h as h goes to 0?

1. Nov 29, 2013

### student34

1. The problem statement, all variables and given/known data

What is the limit of (1-1)/h as h goes to 0?

2. Relevant equations

f'(x) = Lim(h→0) (f(x+h) - f(x))/h

3. The attempt at a solution

The answer is obviously 0, but I cannot figure out exactly how they got that answer.

2. Nov 29, 2013

### lurflurf

(1-1)/h=0
for all nonzero h
therefore
$$\lim_{h\rightarrow 0}\frac{1-1}{h}=\lim_{h\rightarrow 0}0$$

3. Nov 29, 2013

### Student100

I think I stole your name, or you stole mine. ;)

Maybe it would also be helpful if you looked at this way:

When thinking about limits, h is never actually 0. We're interested in what happens as h gets so close to zero we basically treat it as such. Since what you wrote is equivalent to saying $$\frac {0}{.000000....1}$$ you can conclude that this limit is 0.

If on the other hand you had wrote:

$$\lim_{h\rightarrow 0}\frac{(1+h)-1}{h}= 1$$

Would be the correct interpretation of the limit.

Last edited: Nov 30, 2013
4. Dec 10, 2013

### student34

Isn't the limit 0?

5. Dec 10, 2013

### lurflurf

The other student was comparing two different limits.

$$\lim_{h \rightarrow 0} \frac{1-1}{h}=0 \\ \lim_{h \rightarrow 0} \frac{1+h-1}{h}=1 \\ \text{both special cases of}\\ \lim_{h \rightarrow 0} \frac{1+a \, h-1}{h}=a$$