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## Zenmo one year ago Small Question (formula related). Finding equation of tangent line at the given point.

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1. Zenmo

Here are the formulas: $\lim_{x \rightarrow a}\frac{ f(x)-f(a) }{ x-a }$ (#1)

2. Zenmo

$\lim_{h \rightarrow 0}\frac{ f(a+h)-f(a) }{ h }$ (#2)

3. Zenmo

Two example problems: (A) $y=x^3-3x+1; Point (3,19)$ (B) $y=\sqrt{x} ; Point (25,5)$

4. Zenmo

For problem (A), it uses formula (#2). For problem (B), it uses formula (#1). How do I know what formulas to use, so I won't use formula #1 on example A and formula #2 on example B?

5. freckles

you can use either 1 or 2

6. anonymous

yea..

7. Zenmo

Ah I see, so either can work.

8. anonymous

i used to get familiar with equation 2 for all the problems when i started calc

9. anonymous

equation 2 is derived by equation 1. let h=x-a

10. anonymous

its just a slightly more neater way

11. anonymous

I'm sure freckles will give a more clear view

12. freckles

example: $\lim_{x \rightarrow 3} \frac{f(x)-f(3)}{x-3} \\ =\lim_{x \rightarrow 3} \frac{(x^3-3x+1)-(3^3-3(3)+1)}{x-3} \\ =\lim_{x \rightarrow 3} \frac{(x^3-3^3)+(-3x+3(3))+(1-1)}{x-3 } \\ =\lim_{x \rightarrow 3} \frac{(x-3)(x^2+3x+3^2)-3(x-3)}{x-3} \\ \text{ divide \top and bottom by } x-3$ $\lim_{x \rightarrow 3} \frac{(x^2+3x+3^2)-3}{1} =\lim_{x \rightarrow 3}(x^2+3x+3^2-3) \\ \text{ plug in 3} \\ 3^2+3(3)+3^2-3=3(3^2)-3=27-3=24$ I hope it is obvious I used the difference of cubes formula above

13. freckles

oops and that example was for f(x)=x^3-3x+1 at x=3 :

14. anonymous

$\lim_{x \rightarrow a}\frac{ f(x)-f(a) }{ x-a }$ let some variable $h=x-a$ Substitute 'h' into the first equation (note x=h+a) $\lim_{h-a \rightarrow a}\frac{ f(a+h)-f(a) }{ h }$ $\lim_{h \rightarrow 0}\frac{ f(a+h)-f(a) }{ h }$

15. anonymous

both are inter-related

16. anonymous

oops

17. anonymous

i think i made a mistake

18. Zenmo

second, I need "checking" on this problem.

19. anonymous

should be h+a==>a

20. anonymous

$\lim_{h+a \rightarrow a}\frac{ f(a+h)-f(a) }{ h }$

21. freckles

$\lim_{h \rightarrow 0} \frac{\sqrt{25+h}-\sqrt{25}}{h} \\ \text{ multiply top and bottom by top's conjugate } \\ \lim_{h \rightarrow 0} \frac{\sqrt{25+h}-\sqrt{25}}{h} \cdot \frac{\sqrt{25+h}+\sqrt{25}}{\sqrt{25+h}+\sqrt{25}} \\ \lim_{h \rightarrow 0} \frac{(25+h)-(25)}{h(\sqrt{25+h}+\sqrt{25})} \\ \lim_{h \rightarrow 0} \frac{h}{h(\sqrt{25+h}+\sqrt{25})} \\ \frac{h}{h}=1 \text{ so } \\ \lim_{h \rightarrow 0} \frac{1}{\sqrt{25+h}+\sqrt{25}}=\frac{1}{\sqrt{25}+\sqrt{25}} =\frac{1}{2 \sqrt{25}} =\frac{1}{2(5)}=\frac{1}{10}$ and this is using the other route for the other function you had there at x=25

22. freckles

though yeah to prove they are the same definition and you can use a substitution like @chris00 is doing above

23. Zenmo

If I use formula #2 for example B. $\lim_{h \rightarrow 0}\frac{ f(25+h)-f(25) }{ h }$ =$\lim_{h \rightarrow 0}\frac{ \sqrt{25+h}-\sqrt{25} }{ h }$ =$\frac{ \sqrt{25+h}-5 }{ h }\times \frac{ \sqrt{25+h}+5 }{\sqrt{25+h} +5 }$ =$\lim_{h \rightarrow 0}\frac{ 1 }{ \sqrt{25+h}+5 }$

24. freckles

yes!!!! replace h with 0 now

25. Zenmo

plug in 0 to get 1/10

26. freckles

yep yep

27. Zenmo

but for some reason, it counts that as wrong answer

28. freckles

what is the question?

29. freckles

exactly

30. Zenmo

31. Zenmo

Question #2

32. freckles

it asked for tangent line

33. freckles

not slope

34. Zenmo

ops nvm

35. freckles

but yes slope is required you just need more info

36. Zenmo

Im derping agn, I got it now, insert into point-slope form to get tangent equation

37. Zenmo

ok, im going on study break now, Gonna take a step outside haha

38. Zenmo

Thanks all! :)

39. freckles

lol good luck

40. freckles

and sorry didn't mean to give you the answer on that one I kind of thought they had an example or something with that one definition already given

41. Zenmo

Oh no problem at all, you been greatly helpful so far.

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