- #36
DMOC
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DMOC said:[tex]\lim_{h \to 0}\frac{f(h)}{h} + \lim_{h \to 0}\frac{2xh}{h}[/tex]
The second part would just cancel out since it's a straight up 2xh/h problem wher ethe h's cancel. 2x remains.
f ' (x) = 7 + 2x
Just wondering, but how did you replace f(x) with 2xh?
Thanks for helping man ... are you a teacher?
But there's just one more thing I'm confused about after this problem.
[tex]\lim_{h \to 0}\frac{f(x+h) - f(x)}{h}[/tex] is the formula for derivatives as h approaches zero.
However, we eventually ended up with what you see in the quotes.
[tex]\lim_{h \to 0}\frac{f(h)}{h} + \lim_{h \to 0}\frac{2xh}{h}[/tex]
I had to substitute h in for y but you said that I wasn't supposed to do it, but that's how we got there ... thanks again for your help.