Basics of Calculus


By: Shreyas Singh

What is calculus?

Calculus comes from Latin for "small stone." It deals with functions and what happens when you break them into very small parts, infinitesimally small. There are two main branches of calculus: differential calculus and integral calculus. Differential calculus deals with derivatives, and integral calculus deals with integrals. The first thing you will need to know is how to evaluate limits for functions.


What is a limit?

Imagine a function f(x), and some value of x, c. Now, imagine that you evaluate the function for values that are extremely close to c, but not equal to c. You could evaluate it for c - 1, then c - 0.5, then c - 0.1, then c - 0.000001, and so on. You could also evaluate it for values like c + 1, c + 0.5, c + 0.1, c + 0.000001, and so on, getting closer and closer to c. The value that f(x) approaches when x approaches c is called the limit of f(x) as x approaches c. Common sense might tell you that the limit of f(x) as x approaches c is simply f(c). However, for some functions, that might not be the case.


For example, let us evaluate the limit of 1/x2 as x approaches 0. Here, f(x) is 1/x2 and c is 0. Let us try plugging in some values for x that are close to 0. If x is 1, f(x) is 1. If x is 1/10, f(x) = 1/(1/10)2 = 1/(1/100), which is 100. If x is 1/100, then f(x) is 1/(1/100)2 = 1/(1/10000), which is 10000. We can see that as x gets closer and closer to 0, f(x) increases boundlessly. If we approach from the other side, when x is less than 0, we get the same result. f(-1) = 1, f(-1/10) = 100, f(-1/100) = 10000. Thus, we can say that the limit as x approaches 0 of 1/x2 is infinity. This is one of the cases where the limit of a function at a certain value for x is not the same as the value of the function there, because the function is not defined at x = 0.

Sometimes the limit of a function at a certain place doesnt exist. For example, let us try to find the limit of 1/x as x approaches 0. It might be tempting to say that the limit is infinity, because if you plug in positive values for x that are close to 0, the function approaches infinity. However, if you plug in negative values that are close to 0, the function will approach negative infinity. Since the limits from the positive and negative sides don't match, we say that the limit doesn't exist.

The notation of limits is as follows:


Limit practice problems:

Evaluate these following limits:

  1. Limit as x approaches 5 of 3x + 7
  2. Limit as t approaches -2 of t3 - 2t
  3. Limit as x approaches 7 of (2x2 - 17x + 21)/x - 7
  4. Limit as x approaches infinity of sin(1/x)
  5. Limit as h approaches 0 of ((2 + h)2 - 4)/h

Solutions

  1. For this problem, you could try plugging in values for the function that are very close to 5 and evaluate the function. However, we know that this function has no holes, jumps, or asymptotes, which means that it is continuous at all points. This means that the limit of this function as x approaches any value is the same as the same thing as the function evaluated at that value. Try evaluating the limit as we did in the last question, and you will find that it is the same as the value of the function at 5, which is 3(5) + 7, or 22.
  2. We can do the same thing that we did for the previous problem for this one. We know that it also has no holes, jumps, or asymptotes, so we can evaluate the limit by just plugging -2 into the function. (-2)3 - 2(-2) = -4.
  3. For this one, it is not that straightforward. We cannot evaluate this function at x = 7, because if we plug in 7, the denominator becomes 0, so the function is not defined at that point. We could plug in values close for 7, but there is an easier way. If we factor the numerator 2x2 - 17x + 21 into (2x - 3)(x - 7), we can cancel out the x - 7 in the denominator with the x - 7 in the numerator, and all we have left is 2x - 3. Now we can plug in 7 for x to get 11.
  4. As x approaches infinity, 1/x approaches 0. because the denominator x gets larger and larger. This means that the quantity in the sine approaches 0. This means that the whole function approaches 0, because sin(0) is 0, and the sine function is continuous at all points.
  5. We can expand the numerator (2 + h)2 - 4 to 4 + 4h + h2 - 4 by expanding the binomial square. The 4 and -4 cancel out, so the numerator becomes 4h + h2
    6. We can now divide both the numerator and the denominator by h, so all that is left is 4 + h. Since the function 4 + h is continuous at all points, we can now plug in 0 for h and get 4 as the limit.

Now that you know the basics of limits, you can move on to learning about derivatives. It is best to start with derivatives, then go to integrals.