# Thread: Proving alternative definition of the difference quitient?

1. ## Proving alternative definition of the difference quitient?

Okay, so I'm not asking for somebody to do this problem for me, I've done the work. I was just wondering if somebody could look over my proof and tell me weather its correct or not. Heres the problem:

PROBLEM:

If:

$f'(x) = \lim_{h \to 0}\frac{f(x+h) - f(x)}{h}$

Then prove that also:

$f'(x) = \lim_{k,h \to 0^{+}} \frac{f(x+h) - f(x-k)}{h+k}$
Now heres my proof, please point out any errors:

$\lim_{k,h \to 0^{+}} \frac{f(x+h) - f(x-k)}{h+k}$

$= \lim_{k,h \to 0^{+}} [\frac{f(x+h) - f(x-k)}{h+k} + \frac{f(x)}{h+k} - \frac{f(x)}{h+k}]$

$= \lim_{k,h \to 0^{+}} \frac{f(x+h) - f(x) + f(x) - f(x-k)}{h+k}$

$= \lim_{k,h \to 0^{+}} (\frac{1}{h+k})(\;f(x+h) - f(x) + f(x) - f(x-k)\" alt=" = \lim_{k,h \to 0^{+}} (\frac{1}{h+k})(\;f(x+h) - f(x) + f(x) - f(x-k)\" />

$= \lim_{k,h \to 0^{+}} (\frac{h}{h+k})(\;\frac{f(x+h) - f(x)}{h} + \frac{f(x) - f(x-k)}{h}\" alt=" = \lim_{k,h \to 0^{+}} (\frac{h}{h+k})(\;\frac{f(x+h) - f(x)}{h} + \frac{f(x) - f(x-k)}{h}\" />

$= \lim_{k,h \to 0^{+}} (\frac{h}{h+k}) \cdot ( \; \lim_{k,h \to 0^{+}}\frac{f(x+h) - f(x)}{h} + \lim_{k,h \to 0^{+}}\frac{f(x) - f(x-k)}{h}\" alt=" = \lim_{k,h \to 0^{+}} (\frac{h}{h+k}) \cdot ( \; \lim_{k,h \to 0^{+}}\frac{f(x+h) - f(x)}{h} + \lim_{k,h \to 0^{+}}\frac{f(x) - f(x-k)}{h}\" />

$= \lim_{k,h \to 0^{+}} (\frac{h}{h+k}) \cdot ( \; f'(x) + \lim_{k,h \to 0^{+}}\frac{f(x) - f(x-k)}{h}\" alt=" = \lim_{k,h \to 0^{+}} (\frac{h}{h+k}) \cdot ( \; f'(x) + \lim_{k,h \to 0^{+}}\frac{f(x) - f(x-k)}{h}\" />

$= \lim_{k,h \to 0^{+}} (\frac{1}{h+k}) \cdot ( \; f'(x) \cdot (\lim_{k,h \to 0^{+}} \; h) + \lim_{k,h \to 0^{+}}f(x) - f(x-k)\" alt=" = \lim_{k,h \to 0^{+}} (\frac{1}{h+k}) \cdot ( \; f'(x) \cdot (\lim_{k,h \to 0^{+}} \; h) + \lim_{k,h \to 0^{+}}f(x) - f(x-k)\" />

$= \lim_{k,h \to 0^{+}} (\frac{k}{h+k}) \cdot ( \; \frac{1}{k}f'(x) \cdot (\lim_{k,h \to 0^{+}} \; h) + \lim_{k,h \to 0^{+}}\frac{f(x) - f(x-k)}{k}\" alt=" = \lim_{k,h \to 0^{+}} (\frac{k}{h+k}) \cdot ( \; \frac{1}{k}f'(x) \cdot (\lim_{k,h \to 0^{+}} \; h) + \lim_{k,h \to 0^{+}}\frac{f(x) - f(x-k)}{k}\" />

$= \lim_{k,h \to 0^{+}} (\frac{k}{h+k}) \cdot ( \; ( \;\lim_{k,h \to 0^{+}}\frac{1}{k} \ \cdot f'(x) \cdot (\lim_{k,h \to 0^{+}} \; h) + f'(x)" alt=" = \lim_{k,h \to 0^{+}} (\frac{k}{h+k}) \cdot ( \; ( \;\lim_{k,h \to 0^{+}}\frac{1}{k} \ \cdot f'(x) \cdot (\lim_{k,h \to 0^{+}} \; h) + f'(x)" />

$= \lim_{k,h \to 0^{+}} (\frac{1}{h+k}) \cdot ( \; f'(x) \cdot (\lim_{k,h \to 0^{+}} \; h) + f'(x) \cdot (\lim_{k,h \to 0^{+}} \; k) \; )$

$= [\lim_{k,h \to 0^{+}} \frac{1}{h+k}] \cdot ( \;\;\lim_{k,h \to 0^{+}} h+ k \;\ \cdot f'(x)" alt=" = [\lim_{k,h \to 0^{+}} \frac{1}{h+k}] \cdot ( \;\;\lim_{k,h \to 0^{+}} h+ k \;\ \cdot f'(x)" />

$= [\lim_{k,h \to 0^{+}} \frac{h+k}{h+k}] \cdot f'(x)$

$= [\lim_{k,h \to 0^{+}} 1] \cdot f'(x)$

$= (1) \cdot [f'(x)]$

$= f'(x)$

Now, is that proof correct?

2. Well, you left out the closing parenthesis here:

But otherwise I cannot see any missteps. However, all you have to do is set k = 0 in the following and you have your result:

If the limit exists for k,h->0+, then it must exist when k is always 0.

3. I have one question - your using the fact that

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

provided the limits exist. But the limit

$\displaystyle \lim_{(h,k)\to(0,0)} \dfrac{h}{h+k}$

doesn't exist!

4. Originally Posted by Danny
...But the limit

$\displaystyle \lim_{(h,k)\to(0,0)} \dfrac{h}{h+k}$

doesn't exist!
All valid manipulations are permissible prior to taking the limit in the final step. Technically, the limit symbol should be present only at the beginning of each line to maintain validity, but that would be all that is necessary.