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NoahsArk

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- Why does the limit process give an exact slope of the tangent line instead of just a very close approximation?

I am taking a summer calculus class now. For years I've been stuck on the question of why the limit process gives us an exact slope of the tangent line instead of just a very close approximation. I don't need to know the reason for this class I'm taking- we are basically just learning rules of derivates and integrals and applying them. However, it bothers me that I still don't understand the reason after this long.

When you take the limit of the rise over run of any function as delta x (which I'll call h) gets closer to 0, you can never make h equal to zero because then you'd have zero in the denominator. Also, if h were zero, you'd have a point and not a line, and a point can't have a slope. Since h can never be zero, how can you ever know the exact slope?

The only explanations I can think of aren't fully satisfying to me, and I'm not sure if they are good ones. One example of an explanation is in taking the slope of ##f(x) = x^2## at the point where x = 2. Using the limit process you would get the slope function ##\frac {4h + h^2} {h}## You can't set h = to zero here but you can get an equivalent function of 4 + h. So the slope at x = 2 is 4. These two slope functions are equivalent but one has a gap and the other one doesn't, so we can say that the slope must be where the gap is since that's the only place where h is zero. This seems a bit hand wavy to me since the simplified slope function of 4 + h is not the exact same function as ##\frac {4h + h^2} {h}##.

Is the explanation more trivial then I'm thinking, and is there just some conceptual misunderstanding I'm having. Or, it it actually complicated and the subject of another class like real analysis, in which case it's ok for me not to have more than an intuitive feeling for why the limit process gives us an exact slope? Thanks

When you take the limit of the rise over run of any function as delta x (which I'll call h) gets closer to 0, you can never make h equal to zero because then you'd have zero in the denominator. Also, if h were zero, you'd have a point and not a line, and a point can't have a slope. Since h can never be zero, how can you ever know the exact slope?

The only explanations I can think of aren't fully satisfying to me, and I'm not sure if they are good ones. One example of an explanation is in taking the slope of ##f(x) = x^2## at the point where x = 2. Using the limit process you would get the slope function ##\frac {4h + h^2} {h}## You can't set h = to zero here but you can get an equivalent function of 4 + h. So the slope at x = 2 is 4. These two slope functions are equivalent but one has a gap and the other one doesn't, so we can say that the slope must be where the gap is since that's the only place where h is zero. This seems a bit hand wavy to me since the simplified slope function of 4 + h is not the exact same function as ##\frac {4h + h^2} {h}##.

Is the explanation more trivial then I'm thinking, and is there just some conceptual misunderstanding I'm having. Or, it it actually complicated and the subject of another class like real analysis, in which case it's ok for me not to have more than an intuitive feeling for why the limit process gives us an exact slope? Thanks