## anonymous one year ago I have two more limit problems, and I should be done for the night. Would you mind helping me?

1. anonymous

2. anonymous

I even have the answer to this problem. I just don't understand the steps on how to solve it.

3. zzr0ck3r

which step do you not understand?

4. anonymous

Steps 1 and 2.

5. zzr0ck3r

$$\dfrac{\frac{1}{a+b}-\frac{1}{a}}{b}=\dfrac{\frac{a}{a}(\frac{1}{a+b})-(\frac{a+b}{a+b})\frac{1}{a}}{b}=\dfrac{\frac{a-(a+b)}{a(a+b)}}{b}=\dfrac{-b}{a(a+b)}*\dfrac{1}{b}=\dfrac{-1}{a(a+b)}$$

6. zzr0ck3r

$$\dfrac{\frac{1}{x+\Delta x}-\frac{1}{x}}{\Delta x}=\dfrac{\frac{x}{x}(\frac{1}{x+\Delta x})-(\frac{x+\Delta x}{x+\Delta x})\frac{1}{x}}{\Delta x}=\dfrac{\frac{x-(x+\Delta x)}{x(x+\Delta x)}}{\Delta x}=\dfrac{-\Delta x}{x(x+\Delta x)}*\dfrac{1}{\Delta x}\\=\dfrac{-1}{x(x+\Delta x)}$$

7. zzr0ck3r

$\frac{a}{b}-\frac{c}{d}=\frac{ad-cb}{bd}$

8. anonymous

Haha yeah just trying to collect and process my thoughts.

9. zzr0ck3r

lol :)

10. anonymous

Could you explain how you got from $\frac{ \frac{ 1 }{ x+x \Delta } -\frac{ 1 }{ x }}{ x \Delta }$ to the part after the = sign?

11. anonymous

(sorry I would type it out, but it may take all night)

12. anonymous

From the first circle to the second circle.

13. anonymous

Well, it's 3am where I'm at, so I think I'll just leave this topic open for now and check it again in the morning.

14. anonymous

You get from the first circle to thee second by multiplying and dividing a fraction for the same quantity, you see? a/a and (a+b)/(a+b), if you multiply and divide for the same quantity (just like adding and subtracting the same quantity) the result doesn't change, but it often becomes easier to find!

15. zzr0ck3r

$$\large \dfrac{\frac{1}{x+\Delta x}-\frac{1}{x}}{\Delta x}$$ Ok I want to get a common denominator which will be $$x(x+\Delta x)$$. On the left hand side of the numerator I need to multiply top and bottom by $$x$$ to make that happen. On the right ahnd side of the numerator I need to multiply top and bottom by $$(x+\Delta x)$$. $$\large =\dfrac{\frac{x}{x}(\frac{1}{x+\Delta x})-(\frac{x+\Delta x}{x+\Delta x})\frac{1}{x}}{\Delta x}=\dfrac{\frac{x-(x+\Delta x)}{x(x+\Delta x)}}{\Delta x}=\dfrac{-\Delta x}{x(x+\Delta x)}*\dfrac{1}{\Delta x}\\ \\ \large =\dfrac{-1}{x(x+\Delta x)}$$

16. thomas5267

The step is trying to bring the two fractions together. $\frac{1}{3}-\frac{1}{5}=\frac{1}{3}\frac{5}{5}-\frac{1}{5}\frac{3}{3}=\frac{5-3}{15}=\frac{2}{15}\\ \frac{1}{x+\Delta x}-\frac{1}{x}=\frac{1}{x+\Delta x}\frac{x}{x}-\frac{1}{x}\frac{x+\Delta x}{x+ \Delta x}=\frac{-\Delta x}{x(x+\Delta x)}\\ \frac{x}{x}=\frac{x+\Delta x}{x+\Delta x}=1$

17. zzr0ck3r

Right^^^

18. zzr0ck3r

This is why i wrote $$\dfrac{a}{b}-\dfrac{c}{d}=\dfrac{ad-cb}{bd}$$ ...Same thing :)

19. thomas5267

I really wish the OP has a temporary brain short circuit or else he/she will be having a horrible time dealing with calculus.

20. zzr0ck3r

The more you do it, the more you get used to it... Calculus helps solidify algebra skills because so much of it rests on long tedious algebra.

21. zzr0ck3r

Like day one and the formal definition of the derivative....

22. zzr0ck3r

somewhat formal...

23. thomas5267

That looks like the left hand limit of the derivative of $$\dfrac{1}{x}$$ if you get what I mean.

24. anonymous

@zzr0ck3r Common denominator! Thank you. For some reason (probably lack of sleep) I couldn't tell what was going on. I feel stupid for not being able to tell. Sorry about that. @thomas5267 Yes, it was a temporary brain short circuit. As I stated earlier, it was 3 in the morning when I was trying to do this, so something so simple just easily slipped my mind.