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Limits suck

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Lets say we have the function f(x) = x2 Then how do we calculate a tangent line at say point 4, using limits? If a tangent line is lim(b->a, (f(b)-f(a))/(b-a)) (sorry for the messy parentheses) wouldn''t that juse equal 0/0? Or.....am I missing something with the calculation of limits?

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Samith, if the points were taken exactly where they were it would come out like that. The point with limits is to take points extremely close to the numbers but not at them. That way you arrive at a limit based on observation of where the values are "pushing" towards but not getting there. like at point 4 you could choose a point 4.1 then 4.09, then 4.01 etc.

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I always thought you were supposed to calculate limits by substituting in the variable with the constant number it''s approaching. Anyway, using you''re method, I get that the limit is 8. So the slope of the tangent line at that point is 8. Hmm, this method is crappy. I don''t much like it. I remember you mentioning using the derivative to solve this type of problem, is this the way you would suggest doing it?

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More rigourously:



You use limits to find tangent lines by taking the derivative, which is a function that represents the slope of the line tangent to a function at a point. Shown above is the differentiation of f(x) = x^2, solved at x = 4.

Later,
ZE.

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I'm a little rusty at this, but for yor specific example:

f(x) = x^2

lim (h->0) [f(x+h) - f(x)] / [x + h - x]

= lim (h->0) [(x+h)^2 - x^2] / h

= lim (h->0) (x^2 + 2xh + h^2 - x^2) /h

= lim (h->0) (2xh + h^2) / h

now we just divide both the numerator and the denominator by h (this is legal because the quantity still remains the same - we are essentially multiplying by one)

= lim (h->0) (2xh/h + h^2/h) / (h/h)

= lim (h->0) (2x + h)

since h->0 lim (h->0) (2x + h) = 2x

so at the point x = 4, the derivative (slope) = 8, and just solve for the tangent line (point slope equation or whatever is your favortie form to solve a line)



edit: doh, beaten







[edited by - CrazyMike on October 15, 2003 9:48:27 PM]

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The reason you can't immediately substitute in a constant is because then you're dealing with points that are literally infinitely close. Instead, you have to use limit theory to mathematically determine the behavior of the secant line as it approaches the tangent line (i.e. as your two arbitrary points approach each other unto "infinite closeness.")

[edited by - zealouselixir on October 15, 2003 9:50:43 PM]

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quote:
Original post by Samith
I always thought you were supposed to calculate limits by substituting in the variable with the constant number it''s approaching. Anyway, using you''re method, I get that the limit is 8. So the slope of the tangent line at that point is 8. Hmm, this method is crappy. I don''t much like it. I remember you mentioning using the derivative to solve this type of problem, is this the way you would suggest doing it?


Samith your answer was correct.

Using derivation, the general rule for an equation of this type is:

f''(x) = (n) * x ^(n-1)

so x^2 would be 2x and plugging in point four yields 8 as the slope of the tangent line.

the "''" is read as prime. "f prime of x"

There are other rules for more complex relations, that require things like the chain rule, implicit differentiation, product and quotient rules. You can research them to give you a better understanding on how to deal with a variety of relations.

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quote:
Original post by ZealousElixir
The problem with handing someone the power rule at this point is that they don''t understand why it works.


Yes, I do agree with you. But he asked for it! He can also research on google for "proof+derivative" to see if it makes sense to him. The concept might be elusive to those who haven''t yet reached calculus, but the mechanics is quite easy to implement.

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quote:
Original post by Nervo
Yes, I do agree with you. But he asked for it!

No, he asked (1) how to calculate a tangent via limits and (2) what method experienced people would use. The second question came as a response to the lame method of doing it manually. It would seem that he wants the mathematical basis for calculating the derivative. After all, the power rule is just a shortcut to a derivative.
quote:
Original post by Nervo
He can also research on google for "proof+derivative" to see if it makes sense to him.

No, and that''s exactly the point. It won''t make sense until he practices the fundamental methods on his own, or is taught the steps of the proper way. Of course the mechanics are easy to implement, but only when people hand you the formula, which doesn''t teach anyone anything.

Not flaming, I just have serious issues with circumventing the normal modes of understanding how to problem-solve.

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quote:
Original post by ZealousElixir
quote:
Original post by Nervo
Yes, I do agree with you. But he asked for it!

No, he asked (1) how to calculate a tangent via limits and (2) what method experienced people would use. The second question came as a response to the lame method of doing it manually. It would seem that he wants the mathematical basis for calculating the derivative. After all, the power rule is just a shortcut to a derivative.
quote:
Original post by Nervo
He can also research on google for "proof+derivative" to see if it makes sense to him.

No, and that''s exactly the point. It won''t make sense until he practices the fundamental methods on his own, or is taught the steps of the proper way. Of course the mechanics are easy to implement, but only when people hand you the formula, which doesn''t teach anyone anything.

Not flaming, I just have serious issues with circumventing the normal modes of understanding how to problem-solve.


Well ZealousElixir, I''m not about to try and write out proofs and methods for differentiating for Samith. I don''t see any problem in showing him the mechanics of it if he doesn''t need it for school or something critical. I think you need to calm down.

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Guest Anonymous Poster
the derivative only solves for slope, you still have to solve the equation for mx, which is 8 * 4. So the equation for the tangent line is found by doing:

16 = 32 + b
b = 16 - 32 = -16

so the equation in point slope form of the line tangent to the curve y = x^2 is y = 8x - 16

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the derivative only solves for slope, you still have to solve the equation for b taking into account *mx* (slope multiplied by time) which is 8 * 4. So the equation for the tangent line is found by doing:

16 = 32 + b
b = 16 - 32 = -16

so the equation in point slope form of the line tangent to the curve y = x^2 is y = 8x - 16

EDIT: and I got the 16 because in the original equation 4^2 equals 16 (that is the value of the single point you want to touch)

[edited by - Shadow12345 on October 15, 2003 10:32:47 PM]

[edited by - Shadow12345 on October 15, 2003 10:34:31 PM]

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Hmm, about that n*xn-1 rule, while I haven''t found a proof, I worked it out for f(x) = x3

I don''t see how you would prove it, but I see how it works. It''s really simple, as it can be derived from simple algebra. I don''t know about the proof though, I''m assuming it''s probably a bit over my head. But, whatever, thanks for all the help

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quote:
Original post by Nervo
Well ZealousElixir, I''m not about to try and write out proofs and methods for differentiating for Samith. I don''t see any problem in showing him the mechanics of it if he doesn''t need it for school or something critical. I think you need to calm down.


Uhoh, we''ve devolved into addressing by name and telling the other person to become less anally-retentive. For the love of all things good, I''ll concede. You''re right after all - if it''s not for school, what''s the point of understanding how it works?

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quote:
Original post by ZealousElixir
quote:
Original post by Nervo
Well ZealousElixir, I'm not about to try and write out proofs and methods for differentiating for Samith. I don't see any problem in showing him the mechanics of it if he doesn't need it for school or something critical. I think you need to calm down.


Uhoh, we've devolved into addressing by name and telling the other person to become less anally-retentive. For the love of all things good, I'll concede. You're right after all - if it's not for school, what's the point of understanding how it works?


You appear unable to critically read. I'm not saying there is no point to understanding how it works, yet I am saying that it is not critical for him to understand it at his current position in math, but resources are available on the net for him to give it a try.

[edited by - nervo on October 15, 2003 10:49:46 PM]

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quote:
Original post by Samith
I always thought you were supposed to calculate limits by substituting in the variable with the constant number it''s approaching.

When you can "plug in the value" is refered to as continuity. Since this sounds like a beginning calc class, I''m sure you''ll cover that soon enough

The idea is loosely, if the graph can be drawn without picking up your pen at the point you are taking the limit, then you can plug the x value in and evaluate. If plugging in creates an expression that is undefined (like 0/0), then you need to try and factor something probably.

This probably doesn''t help you with your original question but just keep it in mind: sometimes you can plug in and sometimes you can''t (in calc 1, the times you can''t are usually ridiculously overt )

Hope that helps

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GaulerTheGoat: I''m not in calculus yet, I''m still in FST (Function, Stats, Trig, for those that don''t go to my school or know what this is or something....) I''m just doing this stuff for fun.

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quote:
Original post by Samith
Hmm, about that n*xn-1 rule, while I haven''t found a proof, I worked it out for f(x) = x3

I don''t see how you would prove it, but I see how it works. It''s really simple, as it can be derived from simple algebra. I don''t know about the proof though, I''m assuming it''s probably a bit over my head. But, whatever, thanks for all the help

True.

You have discovered a theorem of calculus. You should be proud. You have joined the ranks of Newton and Leibniz

If your book is any good, it probably has a big list of these things on the front or back cover. The proof is probably in the section about derivatives.

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This reeks of homework, and I'm closing the thread. Please review the forum FAQ (see link below) and in the future at least follow the guidelines when posting anything that is even close to a homework question.

The FAQ also has links to places where homework is the norm.

Forum FAQ

Graham Rhodes
Senior Scientist
Applied Research Associates, Inc.

[edited by - grhodes_at_work on October 16, 2003 12:37:49 PM]

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