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

[student34]

Homework Statement

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

Homework Equations

f'(x) = Lim(h→0) (f(x+h) - f(x))/h

The Attempt at a Solution

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

 

[lurflurf]

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

 

[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.

You'll learn more about limits when you discuss the epsilon/delta definition of a limit.

 

[student34]

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.

You'll learn more about limits when you discuss the epsilon/delta definition of a limit.

Isn't the limit 0?

 

[lurflurf]

The other student was comparing two different limits.

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