Well, I know that the derivative of a function $f(x)$is defined this way:
$$\frac{df(x)}{dx} = \lim_{\Delta x\to 0}\frac{f(x+\Delta x) - f(x)}{\Delta x}$$
And it's pretty clear that the expression inside the limit will approach the tangent line at a given point. I know that this is the definition of derivative. However, we can't define this to be equals the tangent line at a given point. So how do we know that this limit will in fact be equal the slope of the function?
3 Answers
$\begingroup$It will be arbitrarily close to it. To get rid of ambiguities, we need to define derivative in a different way: Suppose $\epsilon > 0 $ If we can find a $\delta > 0$ such that $| \Delta x | < \delta$
$$ \text{Then it follows that} \; \; \; | \frac{f(x + \Delta x) - f(x)}{\Delta x} - L | < \epsilon$$
And we call $L = \frac{df}{dx} $
$\endgroup$ 1 $\begingroup$The slope of a curve at a point on it's graph, is the gradient of the tangent at that point. Now if you draw the graph of $f$ you will see that the expression inside the limit approaches the gradient of the tangent at the point $x$ when $\Delta x$ tends to zero.
The derivative at a point of a function is a general concept that is defined in a abstract way without considering the graph of the function. It so happens that the geometrical interpretation of the derivate at a point is the gradient of the tangent drawn at that graph.
$\endgroup$ $\begingroup$If I understand your concern correctly, you want to know how the definition of the derivative relates to the intuitive notion of what a tangent is.
Let's assume that we are analysing a twice continuously differentiable function $f(x)$ at the point $x=a$ and that $f''(a)$ is positive.
Intuition tells us that a line $\ell(x)$ that is tangent to the graph of $f$ at the given point $a$ should be such that
$\ell(a) = f(a)$ and
It somehow "sits" on the graph of $f$, that is, it stays on top or below the graph in a neighborhood of $a$ (This may not be true if $f''(a)$ is $0$, see below).
Now, take $\ell(x) = f(a) + f'(a)(x-a)$. We will show that it meets the required conditions. Condition 1. is immediate from the definition.
For condition 2., we know that, by continuity, there should be a neighborhood $V$ of $a$ where $f''(a)>0$. Moreover, $f'(x)$ is increasing in $V$. The line $\ell$ in this case sits below the graph. Indeed, if $x >a$ lies in the neighborhood $V$, $$\begin{align} & f(x) - \ell(x) \ge 0\\ \iff & f(x) - f(a) - f'(a)(x-a) \ge 0 \\ \iff & \frac{f(x)- f(a)}{(x-a)} - f'(a) \ge 0 \\ \iff & f'(c) - f'(a) \ge 0 \\ \end{align}$$ by the Mean Value Theorem, with $c$ lying in the interval $[a,x]$. The last inequality is true because $f'$ is increasing in V. A similar argument shows that $f(x) \ge \ell(x)$ for an $x$ in $V$ with $x<a$.
If $f''(a)$ is negative, you can get the same result, but this time with $\ell$ lying on top of the graph of $f$.
The case $f''(a)=0$ is somewhat different, as you may have something like this:
