MA005: Calculus I

If we have two points on the graph of a function, $(x, f(x))$ and $(x+h, f(x+h))$, then $\Delta y=f(x+h)-f(x)$ and $\Delta x=$ $(x+h)-(x)=h$ so the slope of the secant line through those points is $m_{\text {secant }}=\frac{\Delta y}{\Delta x}$ and the slope of the line tangent to the graph of $f$ at the point $(x, f(x))$ is, by definition,

$\mathrm{m}_{\text {tangent }}=\lim\limits_{\Delta x \rightarrow 0}\{$ slope of the secant line $\}=\lim\limits_{\Delta x \rightarrow 0} \frac{\Delta y}{\Delta x}=\lim\limits_{h \rightarrow 0} \frac{f(x+h)-f(x)}{h}.$

Example 3: Give a geometric interpretation for the following limits and estimate their values for the function in Fig. 5:

(a) $\lim\limits_{h \rightarrow 0} \frac{f(1+h)-f(1)}{h}$    (b) $\lim\limits_{h \rightarrow 0} \frac{f(2+h)-f(2)}{h}$

Solution: Part (a) represents the slope of the line tangent to the graph of $\mathrm{f}(\mathrm{x})$ at the point $(1, \mathrm{f}(1))$ so

$\lim\limits_{h \rightarrow 0} \frac{f(1+h)-f(1)}{h} \approx 1$. Part (b) represents the slope of the line tangent to the graph of $\mathrm{f}(\mathrm{x})$ at the point $(2, \mathrm{f}(2))$ so $\lim\limits_{h \rightarrow 0} \frac{f(2+h)-f(2)}{h} \approx-1$.

Practice 4: Give a geometric interpretation for the following limits and estimate their values for the function in Fig. 6:

$\lim\limits_{h \rightarrow 0} \frac{g(1+h)-g(1)}{h} \quad \lim\limits_{h \rightarrow 0} \frac{g(3+h)-g(3)}{h} \quad \lim\limits_{h \rightarrow 0} \frac{g(h)-g(0)}{h}$