What does it mean to evaluate a limit after finding a difference quotient?

[Jaclbl]

I understand how to find a difference quotient, but afterwards it asks me to then evaluate or approximate each limit, is that just by plugging in the given limit or is there another step?

 

[Jameson]

I understand how to find a difference quotient, but afterwards it asks me to then evaluate or approximate each limit, is that just by plugging in the given limit or is there another step?

Hi Jaclbl,

Welcome to MHB! :)

It would be much better to look at a real problem I think, but in general these limits usually require a cancellation because in the denominator you have a $\Delta x$ and the limit is $\Delta x \rightarrow 0$, so some algebraic manipulation is required to hopefully cancel something out.

 

[Jaclbl]

Ah, I don't want to do the exact problem, because it is for homework, but I'll send one of the ones that is like it, for an example.
lim F(x+h)-f(x)
h-0 h

f(x) = 4 - 2x -x²

 

[Jameson]

Ah, I don't want to do the exact problem, because it is for homework, but I'll send one of the ones that is like it, for an example.
lim F(x+h)-f(x)
h-0 h

f(x) = 4 - 2x -x²

Ok, sounds good. What are $f(x+h)$ and $f(x)$ for this problem?

 

[soroban]

Hello, Jaclbl!

$$f(x) \:=\:4-2x-x^2$$

Find: $$\;\lim_{h\to0}\frac{f(x+h) - f(x)}{h}$$

I have taught my students to make four steps.

(1) Find $$f(x+h)$$; replace $$x$$ with $$x+h$$, and simplify.

(2) $$f(x+h) - f(x)$$: subtract $$f(x)$$, and simplify.

(3) $$\frac{f(x+h)-f(x)}{h}$$: divide by $$h$$ . . . factor and reduce.

(4) $$\lim_{h\to0}\frac{f(x+h)-f(x)}{h}$$: take limit as $$h\to 0.$$We are given: $$\:f(x) \:=\:4 - 2x - x^2$$

$$(1)\;f(x+h) \:=\:4-2(x+h) - (x+h)^2 \:=\:4 - 2x - 2h - x^- 2xh - h^2$$

$$(2)\;f(x+h)- f(x)$$
. . . $$=\: (4-2x-2h - x^2 - 2xh - h^2)-(4-2x-x^2) \:=\:\text{-}2h -2xh - h^2$$

$$(3)\;\frac{f(x+h)-f(x)}{h} \;=\;\frac{\text{-}2h-2xh-h^2}{h} \;=\;\frac{h(\text{-}2-2x-h)}{h} \;=\;\text{-}2-2x-h$$

$$(4)\;\lim_{h\to0}\frac{f(x+h)-f(x)}{h} \;=\;\lim_{h\to0}(\text{-}2-2x-h) \;=\;\text{-}2-2x$$Therefore, the derivative of $$f(x)\:=\:4-2x-x^2$$ is: $$\;f'(x) \:=\:-2-2x$$