# Thread: [SOLVED] Prove two definitions of derivative is equivalent

1. ## [SOLVED] Prove two definitions of derivative is equivalent

Hi, everyone! I have a minor problem proving the two following definitions of the derivative are equivalent:

f is differentiable iff $\lim_{x->x_{0}}\frac{f(x)-f(x_{0})}{x-x_{0}}$ exists
and
f is differentiable iff $\lim_{t->0}\frac{f(x_{0}+t)-f(x_{0})}{t}$ exists.
I know that i can let $t=x - x_{0}$, so that $x=x_{0}+t$ and $\frac{f(x) - f(x_{0})}{x - x_{0}}=\frac{f(x_{0} + t) - f(x_{0})}{t}$, but then how can i explain the limits part, when in first case i am taking limit as x approaches $x_{0}$ and in second as t approaches 0?

2. First, be aware that I not really sure that I understand you difficulty here.
That said, surely it is clear to you that this is true:
$\lim _{x \to x_0 } \left( {x - x_0 } \right) = 0\;\& \,\lim _{t \to 0} t = 0.$

Now tell us what you don't understand.

3. Yea, i'm aware of that, but i'm not sure how I can apply that knowledge to my problem, since as far as I am concerned i'm not taking limit of t as t goes to 0, but rather limit of difference quotient as t goes to 0, where difference quotient need not be t. I guess what I really wanted to know is whether $\lim_{x-x_0\to 0}(Y)=\lim_{x \to x_0}(Y)$ is generally true for any $Y$. Correct me if i'm totally off the track.

4. Originally Posted by vanishingpoint
Hi, everyone! I have a minor problem proving the two following definitions of the derivative are equivalent:

f is differentiable iff $\lim_{x->x_{0}}\frac{f(x)-f(x_{0})}{x-x_{0}}$ exists
and
f is differentiable iff $\lim_{t->0}\frac{f(x_{0}+t)-f(x_{0})}{t}$ exists.
I know that i can let $t=x - x_{0}$, so that $x=x_{0}+t$ and $\frac{f(x) - f(x_{0})}{x - x_{0}}=\frac{f(x_{0} + t) - f(x_{0})}{t}$, but then how can i explain the limits part, when in first case i am taking limit as x approaches $x_{0}$ and in second as t approaches 0?

Since you defined $t=x-x_0$ , it follows at once that if $x\rightarrow x_0$ , then $t\rightarrow 0$, and the other way around. This is EXACTLy the reason why both limits above are equivalent.