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Limits are a very important tool in calculus and they are used to determine whether a function will be increasing or decreasing at a given point. They can also be used to determine whether the graph of a function is horizontal or vertical at a given point, or whether it crosses the x-axis or y-axis at that point.
The product rule, power law and quotient rule are three important rules used in calculus.
Now, we will review some of the techniques for finding limits of functions. These are called limit laws. Suppose that f(x) and g(x) both approach c as x approaches c, but f(x) does so with a limit equal to L and g(x) does so with a limit equal to M. Note that the infinity symbol or minus infinity symbol are not considered as real numbers.
The Limit Laws state that if we try to compute the sum of f and g, the limit when x tends to c, we just add the corresponding limits, L with M.
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In a similar fashion, we can also do subtraction.
The limit, as x approaches c, of (f - g) is equal to L - M. In a similar way, if we multiply a fixed number k to a function f, we can compute the limit of this new function by simply multiplying k times L.
We can even multiply two functions, f and g, when x tends to c.
The limit of the product of f and g is simply equal to L times M.
If we do division, f divided by g, we just divide L by M provided that M is not equal to zero. If M equals zero, we will have the expression L over zero which is meaningless.
In addition, if we look at the power of f(x), say f(x) square or third power, forth power, we can compute the limit of this situation by raising L to the same power. L being the limit of f(x) when x tends to c.
Instead of doing power, you can take the nth root, meaning the inverse of a number raised to the power of n.
In the same way, you can apply these Limit Laws to compute limits of functions. For instance, let's consider the following example.
We will consider the function seven x to the power -3, minus one, raised to the one-third power as x tends to 2. First of all, we can take the limit inside the one over three root.
This is the Power Law, where you have to compute the limit of the expression inside.
First, we break it up into two separate pieces by using the first law of arithmetic that we call the difference rule. Then we compute each piece separately.
The first one here, you just put x tends to two and get seven two and minus three. This is a constant function, with a limit of one. Here we make use of already the Product Rule, Power Law and also the Quotient Rule. Finally we just simplify this to get the final answer which is minus one half.
Now recall that when we're trying to compute the ratio f over g as x approaches c, we get the answer L/M provided M is not equal to zero.
But in most cases, when we apply this equation, it will be in a situation where M equals zero. So how are we going to deal with this? Now look at this example:
We are going to compute the limit of this fraction. The numerator is given by x squared minus three x plus two, while the denominator is x squared minus four. When x tends to two, we want to look at this ratio.
If we put x equal to two in the numerator, we get zero; in the denominator, we also get zero. We cannot apply the Quotient Rule directly. We can study this function by writing it as a ratio.
The graph of this function is actually the blue line with one single hole, corresponding to x equals to two. The reason why we have a hole here is that when x equals to two, this function is undefined so the graph does not has any value at x equals to two; we put the hole here.
To compute the limit in this situation, we can factorise the numerator into x minus one times x minus two. Similarly, factorise the denominator and we get x plus two times x minus two.
Now, we can cancel the x-squared term in both the numerator and denominator. We know that x is not equal to two, so x-squared is nonzero. After canceling it out, we get a very simple fraction, which we can then take the limit as both numerator and denominator extend to two.
The numerator becomes one, and the denominator becomes four. The final answer is one over four.
We have canceled the common factor x minus two, which results in this equation: x minus one over x plus two.
The graph shown on the right-hand side here is exactly the same as the one on the left-hand side, except now it does not contain a hole at x = 2 because the function itself is defined at x = 2.
In summary, f(x) and g(x) are not the same function. They differ only at x = 2, but by looking at g(x) instead of f(x), we can compute the behavior of f(x) as x approaches 2. Therefore, we get an answer of 1/4. Now we come to another very useful technique called the Sandwich Theorem.
Suppose we have three functions, g, f and h in this order. For all x in an open interval containing the point c, g is less than or equal to f which is also less than or equal to h. The aim is to compute the limit of f when x tends to c but we do not know what this limit is. This may be because it is a complicated function.
However, if we know that f(x) when x tends to c, and h(x) when x tends to c both have the same limit L, then we can conclude g(x), which is sandwiched between f and h, also has the same limit L as well.
Example.
I have a function f(x) that I do not know precisely but I know that f(x) is always less than or equal to five times x squared. So the function is shown in blue.
The five-times-squared expression is the green one, which stays above the blue one. The negative two-times-squared expression is denoted by this red one, and it always stays
below the blue one. So in this situation, you can see that the blue one lies between the green one and the red one.
This is why we say that f(x) is sandwiched between the green one and the red one. We want to find the behavior of f(x) when x tends to zero, but we know that when x tends to zero, the red line comes very close to value zero.
Similarly, the green one is almost as close to zero as is the blue one. Since both are between the red and green values, they must also be very close to zero.
According to the Sandwich Theorem, which we call the Sandwich Theorem, it can be concluded that f(x) when x tends to zero also has the value zero.
For those who are interested in seeing the proof that f(x) is between g and h, here it is.
If we subtract L from all of them, g(x) minus L is less than or equal to f(x) minus L less than or equal to h(x) minus L.
When x is close to c, h(x) - L is small and g(x) - L is also small. Therefore if you look at the distance between f(x) and L, it is less than or equal to the distance between h and L if f(x) - L is positive and equal to g(x) - L if f(x) - L is negative.
We can conclude that f(x) goes to L as x goes close to c since both the numerator and denominator of the fraction get very small as x approaches c. This idea is the basis for proving the Sandwich Theorem.
Limit of a Function and Limit Laws (Part II). (2023, Aug 04). Retrieved from https://studymoose.com/limit-of-a-function-and-limit-laws-part-ii-essay
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