Limits suck
Author
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?
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.
Author
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?
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.
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.
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]
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]
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]
[edited by - zealouselixir on October 15, 2003 9:50:43 PM]
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.
The problem with handing someone the power rule at this point is that they don''t understand why it works.
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.
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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