FFR - Limits and Slopes

Flexibility of Limits

In the previous section, we briefly touched on how limits can be used to define actual functions in Gallery 2.3 with f(x)=e^x. Limits are actually even more flexible than that. A limit can be used to find the slope of a line tangent to any point on a curve! Before we go into that, however, recall how to find the slope, m:​
$m=\frac{y_2-y_1}{x_2-x_1}=\frac{f(x_2)-f(x_1)}{x_2-x_1}$ Plugging in two function values corresponding to two x-values in the above graph would give the slope of the secant line going through those two points of the function. Imagine what might happen as these two points came really close to each other, and the secant line began to go through a smaller and smaller section of the graph, until the two points it went through were so close that it looked like it only went through one point. This is shown on the next page.

Other Methods of Solving Limits

Ever since computers became widely used, it has been possible to easily solve limits through simply being able to read and understand graphs. However, the programmers of these computers had to understand how these limits worked. Besides solving limits graphically, it is also possible to solve them using approximations, by calculating how the function behaves near the limit. If only you could bring laptops to your Calculus exams!

Gallery 2.4.1 explains visually the concept behind secant and tangent lines. In the first image of the gallery, notice how this is the traditional way of looking at the slope of a secant line going through two points in a graph. What would happen as x approaches a?

The second image in the gallery demonstrates a slightly different way of looking at the same problem of the slope of a secant line going through two points of a known function. By replacing (x-a) and x approaching a with a single term, h, approaching zero, we can make the entire equation easier to solve as a limit.

The third and final image demonstrates this visually. What would happen as h approaches zero? We would get the slope of a tangent line to the point of interest, in this case, (a, f(a)).

What does this mean?

Absolutely nothing... No, it means something very interesting. It means that by using limit properties, we can calculate the slope of a tangent line at any point on a graph. Think about it, (a, f(a)) is an arbitrary point we have selected of our own free will. What if we replace (a, f(a)) with (x,f(x))? In this case, x is a variable. If we varied our point of interest, we are still picking another point very close to it as h approaches 0, so we still get the slope of a tangent line at that point! The slope of the tangent line at a in the last slide can be calculated by the limit $\underset{h\to 0}{lim}\frac{f(a+h)-f(a)}{h}$ . In this case, this equation would give a negative answer, and that is to be expected since the slope is negative! Now if we replace a constant a with the variable x, the general slope of a line that is tangent to any point on a graph becomes:

$\underset{h\to 0}{lim}\frac{f(x+h)-f(x)}{h}$

The equation above is a very special limit in calculus. It is so special that half of the rest of this book is based on that limit, and even has its own name, a derivative.

Note: Derivative Rules - Although a derivative is based on limits, many of the laws do not carry over. The only laws we can apply to derivatives that we can also apply to limits are addition/subtraction and coefficient laws. When finding general derivatives of one function multiplied/divided by another function, or composed into another function, derivatives have their own rules. These will be covered in later chapters.

Example Problem 2.4.1.

Find the derivative function of $f(x)=x^2$ using limits.

Chapter 2 Problems

1. Find ε in terms of δ for the function $f(x)=4x+1$

2. Solve the limit $\underset{x\to \mathrm{\infty }}{lim}\frac{\frac{4x^3-3x}{x^3}}{\frac{5x^2+2x+5}{x^2}}$

3. Find $\underset{x\to 4}{lim}f(x)=\{\begin{array}{rlr}\sqrt{x},& & ifx\ne 4\\ 4.& & ifx=4\end{array}$
4. Solve $\underset{x\to 0}{lim}\frac{\cos (x)-1}{\sin (x)}$

5. Simplify and solve $\underset{x\to 0}{lim}\frac{2(1-{\cos }^{2}x)}{2\sin x+\sin (2x)}$
6. Solve the limit $\underset{x\to 0}{lim}\frac{\cos x\left(\cos x-1\right)+\sin x}{x^2}$

7. Find $\underset{x\to 0}{lim}\frac{\sqrt{4-x}-2}{\sqrt{x+1}-1}$
8. Find $\underset{x\to 1}{lim}\frac{\sqrt{x^2+3}-2}{\sqrt{x^2+8}-3}$

9. What is $\underset{x\to 2}{lim}\frac{\sqrt{x^2+5}-3}{\sqrt{x^2+14}-4}$ ?
10. Prove that the derivative of a constant is 0 at all points on its graph (Hint: You must identify the difference between “approaches” and “equals”).