Question

Limit Problem, See comments for equation:

Answer 1

Use the limit definition to find the slope of the function: \[f(x)= 2/(x-1)\]

Answer 2

\[f'(x)=\lim_{h \rightarrow 0}\frac{f(x+h)-f(x)}{h}=\lim_{h \rightarrow 0}\frac{\frac{2}{x+h-1}-\frac{2}{x-1}}{h}\]

Answer 3

did you get this for the first step?

Answer 4

\[=\lim_{h \rightarrow 0}\frac{\frac{2(x-1)}{(x+h-1)(x-1)}-\frac{2(x+h-1)}{(x-1)(x+h-1)}}{h}=\lim_{h \rightarrow 0}\frac{\frac{2(x-1)-2(x+h-1)}{(x+h-1)(x-1)}}{h}\]

Answer 5

\[=\lim_{h \rightarrow 0}\frac{\frac{2x-2-2x-2h+2}{(x+h-1)(x-1)}}{h} \cdot \frac{\frac{1}{h}}{\frac{1}{h}}\]

Answer 6

\[\lim_{h \rightarrow 0}\frac{-2h}{(x+h-1)(x-1)} \cdot \frac{1}{h}\]

Answer 7

\[\lim_{h \rightarrow 0}\frac{-2}{(x+h-1)(x-1)}=\frac{-2}{(x+0-1)(x-1)}=\frac{-2}{(x-1)(x-1)}=\frac{-2}{(x-1)^2}\]

Answer 8

Alright. Not the way I went with it, what so ever. I'm going to have to spend some time understanding your steps. Again, myininaya, you've helped me out, thank you very much!

Answer 9

I have another way in my mind i can put that down if you like

Answer 10

\[\lim_{h \rightarrow 0}\frac{\frac{2}{x+h-1}-\frac{2}{x-1}}{h} \cdot \frac{(x+h-1)(x-1)}{(x+h-1)(x-1)}\]
\[\lim_{h \rightarrow 0}\frac{2(x-1)-2(x+h-1)}{h(x+h-1)(x-1)}\]

Answer 11

\[\lim_{h \rightarrow 0}\frac{2x-2-2x-2h+2}{h(x+h-1)(x-1)}\]

Answer 12

\[\lim_{h \rightarrow 0}\frac{-2h}{h(x+h-1)(x-1)}=\lim_{h \rightarrow 0}\frac{-2}{(x+h-1)(x-1)}\]

Answer 13

and i did this part above

Answer 14

instead of combining the fractions like I did in the first way, I could have first just multiply by the factors of numerator's denominator like i did in the 2nd way

Answer 15

denominators*

Answer 16

Again, thank you for shedding some light on this. My algebra is a little rusty, and these problems really exploit it. I will have to go back and work on that.