Definition of a Limit

[Matthew Ko]

I have a feeling that F(x)=sec x. Then, to solve for f'(x), I'd take the second derivative of sec x. However, how can I prove so? Even if I change a to t, would "the limit of F(x) as x approaches to t" accounted as same as "the limit of f(x) as t approaches to x"?

Thank you for your time in reading and helping with my question.

 

[Dr.Peterson]

It's hard to distinguish your letters (F, f, t, x). I see no reason to define a new name F, but you're right that f(x) is the derivative of sec(x), so you can just find that derivative, and then take its derivative in turn.

Your version of the definition gives F'(a), using x as a temporary variable. By not stating this, you allowed yourself to be confused. Just replace x with t as the temporary variable, and a with x, and you have the limit you were given.

 

[Matthew Ko]

Thank you for your reply. I attached a typed version of my handwriting above.

I can see how if I do as you advised, the two functions align up neatly. However, doesn't x has to be the temporary variable (not t) because x is the function's input variable (that is, the function is f(x) not f(t))?

 

[Dr.Peterson]

I can see how if I do as you advised, the two functions align up neatly. However, doesn't x has to be the temporary variable (not t) because x is the function's input variable (that is, the function is f(x) not f(t))?

I'm not sure what you mean. I hope you know that [MATH]f[/MATH] is the same function regardless of what you call the input variable.

In the definition as you state it, [MATH]F'(a) = \lim_{x\to a}\frac{F(x)-F(a)}{x-a}[/MATH], you are finding the derivative at a specific point [MATH]a[/MATH]. The variable [MATH]a[/MATH] is fixed, and [MATH]x[/MATH] is moving toward it.

In the expression you are comparing to this, [MATH]F'(x) = \lim_{t\to x}\frac{F(t)-F(x)}{t-x}[/MATH], you are finding the derivative at a specific point [MATH]x[/MATH], by letting a variable [MATH]t[/MATH] move toward [MATH]x[/MATH]. These match up perfectly.

It is common for students to struggle a bit with this: As we take the limit, the value of [MATH]x[/MATH] is fixed, because we are finding the rate of change at a given point. But [MATH]x[/MATH] can take any value, so we end up with the derivative being a function of [MATH]x[/MATH]. This is why I said it's important to write not just your [MATH]m_{F(x)}[/MATH], which is a notation I don't think I've ever seen before, but [MATH]F'(a)[/MATH], which makes it clear where you are finding the derivative. In your notation, [MATH]x[/MATH] plays an ambiguous role; if I were to use such a notation, I would just say [MATH]m_F[/MATH], indicating that this is the slope of function [MATH]F[/MATH]. It is not the slope at some fixed value called [MATH]x[/MATH], but at [MATH]x=a[/MATH]!

 

[pka]

View attachment 22230

I have a feeling that F(x)=sec x. Then, to solve for f'(x), I'd take the second derivative of sec x. However, how can I prove so? Even if I change a to t, would "the limit of F(x) as x approaches to t" accounted as same as "the limit of f(x) as t approaches to x"?

@Matthew Ko, your notation as posted is an absolute abomination.
You have \(f(t)=\mathop {\lim }\limits_{t \to x} \dfrac{{\sec (t) - \sec (x)}}{{t - x}}\)
Are you really that challenged when it comes to mathematical notation?
BTW: Your handwriting can have much improving.
Look, everyone can read his/her own handwriting. That is not the problem.
BUT you are asking us for free professional help. So why not post readable posts as a simple courtesy?
You should learn to use LaTeX cording if you do not improve the readability of your posts.

 

[Matthew Ko]

Thank you for your help and replies. Dr. Peterson, your explanation helped a lot and I think I now understand. I also apologize to both of you for the inconvenience by causing readability issues. I will make sure I post readable materials next time.

 

[HallsofIvy]

Let h= t- x. Then t= x+ h and, as t goes to x, h goes to 0. The limit becomes \(\displaystyle \lim_{h\to 0}\frac{sec(x+h)- sec(x)}{h}\).

In other words the limit is the derivative of sec(x).

If you already know that derivative, you are done. If not then perhaps this is an exercise in determining that derivative.

To do that, use the fact that \(\displaystyle sec(x)= \frac{1}{sin(x)}\) so the limit can be written \(\displaystyle \lim_{h\to 0}\frac{1}{sin(x+h)}- \frac{1}{sin(x)}{h}\).

Simplify the numerator by gettng the common denominator sin(x+h)sin(x)- the numerator is
\(\displaystyle \frac{sin(x)- sin(x+h)}{sin(x+h)sin(x)}\)
so the limit can be written
\(\displaystyle \lim_{h\to 0}\frac{sin(x)- sin(x+ h)}{h sin(x)sin(x+ h)}= \left(\lim_{h\to 0}\frac{1}{sin(x)sin(x+h)}\right)\left(-\lim_{h\to 0}\frac{sin(x+h)- sin(x)}{h}\right)\).

The first limit is \(\displaystyle frac{1}{sin^2(x)}\) while the second is the negative of the derivative of sine, -cos(x).

The limit is \(\displaystyle -\frac{cos(x)}{sin^2(x)}=-\frac{cos(x)}{sin(x)}\frac{1}{sin(x)}= -cot(x)csc(x)\).[/tex][/tex]