Why are these two derivative functions equivalent?

Question

Question one:

The derivative of function $f$ at the point where $x=t$ is defined by the following two equivalent limit expressions: $$\lim_{x\to t} \frac{f(x)-f(t)}{x-t}$$ $$\lim_{h\to0} \frac{f(t+h)-f(t)}{h}$$

I understand the first expression - same as basic algebra (change in $y$ over change in $x$).

Yet I don't get how the second expression is equivalent really. I'm just starting with derivatives so yeah if someone could explain it to me, that would be lovely.

Question two:

Lets substitute $f(x)=\frac{1}{x}$ and $t=1$ into the second expression to find the appropriate limit expression for $f'(1)$. This returns end result: $$\lim_{h\to0} \frac{\frac{1}{1+h}-1}{h}$$

So even if I would understand how on the first two questions the expressions are equivalent, how is a function of $f(x)=\frac{1}{x}$ and $t=1$ have anything to do with that end result...

Thanks in advance for any clearance on the subject.

Answer 1

$Q1$: Ok, suppose you have points $(t, y1), (x,y2)$, without loss of generality, let $x>t$, since we want distinct points on the function. Then let $h$ denote this horizontal distance, then the two points translate to $(t, f(t)), (t+h,f(t+h))$, the slope of the secant line between the two points is $$ \frac{f(t+h)-f(t)}{h}$$Note the derivative is the slope of the tangent line at one given point, and hence if we want the slope at $x=t$, then we want the secant line to move closer, infinitely close so that it becomes a tangent line. Thus, we want to make their horizontal distance infinitely small. Which means we want to limit $h$ to $0$. Hence we have the formula $$\lim_{h\to0} \frac{f(t+h)-f(t)}{h}$$

To understand the second question, I think you can go back to my explanation of the formula for $Q1$, since one of the points started at $x=t$, we would need the y value when calculating slope, which is why it matters what value the function has at that point. $Q2$ is highly dependent on your understanding on $Q1$, hope that my explanation for $Q1$ can help with your understanding with $Q2$.

Answer 2

For question one note that the second expression is obtained by a change of variable

• $h= x-t \to 0$

$$\lim_{x\to t} \frac{f(x)-f(t)}{x-t}\iff \lim_{h\to0} \frac{f(t+h)-f(t)}{h}$$

For the second question just observe that

• $f(1+h)=\frac{1}{1+h}$
• $f(1)=1$

Answer 3

Question one:

The idea behind both expressions is, as you pointed out, change in $f(x)$ over change in $x$.

With the first expression, the inner expression is the change between two points, $x$ and $t$. The overall expression, including the limit, represents the value of this change in $f(x)$ over change in $x$ as $x$ gets closer to $t$.

With the second expression, as someone else mentioned, this is a change of variable. Instead of using two points, $x$ and $t$, let's use one point $t$ and change $x = t + h$; this defines $x$ in terms of the point $t$, +/- some distance from this point which we're calling $h$.

Instead of viewing it as two specific points, $x$ and $t$, and calculating the limit as $x$ approaches $t$, the second expression is just saying: take one point, $t$, and an arbitrary (nonequal) point $t + h$, and calculate the limit as the difference between the two points becomes zero.

It's simply a matter of viewpoint, where expression one represents calculating the limit as one point approaches another, and expression two represents calculating the limit as the difference between the two points approaches zero (still with the same point, $t$, fixed).

Question two:

We are given that: $$f(x) = \frac{1}{x}, t= 1$$ Looking at the second expression, and substituting the above, we need to determine $f(t+h)$ and $f(t)$: $$f(t + h) = \frac{1}{t + h} = \frac{1}{1 + h}$$ $$f(t) = \frac{1}{t} = \frac{1}{1} = 1$$ Then we can plug these into the second expression from Question one: $$f'(1) = \lim_{h\to 0}\frac{f(t+h)-f(t)}{h} = \lim_{h\to 0}\frac{\frac{1}{1 + h}-1}{h}$$