 # Limits, the First Step into Calculus

by | Mar 25, 2020 | Resources

The concept of a limit is the central idea that underlies calculus and is the unifying mechanism that allows for differentials and integrals to be related. Calculus is used to model real-life phenomena in the language of mathematics. Anything that involves a rate of change, as the velocity of your car is the rate of change of distance with respect to the rate of change of time, is found using derivatives. Limits are the basis of the derivative, by finding the instantaneous rate of change.

## Definition of a Limit

The limit is the behavior of a function as we approach a certain value. Let’s start by looking at a particular function

$$f(x) = x^2 + x – 6$$

for values near 2. We can use a table of values that gets really close to 2 from values less than 2, and another that gets really close to 2 from values greater than 2.

From the table, we can see that as x approaches 2, the value of f(x) approaches 0. It would appear from the chart, that if we let x get really close to 2 in either direction that f(x) becomes 0. This is the basic version of how we solve a limit. We use the English phrase “the limit of f(x) as x approaches 0 is equal to 0”.

## The Limit of a Function: Definition

We say

$$\lim_{x\rightarrow a} f(x) = L$$

and

$$\text{“The limit of f(x) as x approaches a equals L”}$$

if when we make the value of x get arbitrarily close to a, the value of f(x) gets arbitrarily close to L.

## Finding Limits by “Direct Injection”

If we are searching for a limit like

$$\lim_{x\rightarrow 5} x^2+x-10$$

we can do what is called “Direct Injection”, in other words we plug in the value of $$x=5$$ into the function we are finding the limit of

$$\lim_{x\rightarrow 5} 5^2+5-10=20$$

then we have discovered that the limit is equal to 20.

Try this method on the following problem:

#### Example

Find the limit

$$\lim_{x\rightarrow 1} \frac{x-1}{x^2-1}$$

#### Solution

If we try direct injection we have a problem:

$$\lim_{x\rightarrow 1} \frac{1-1}{1^2-1} = \frac{0}{0}$$

The problem is that we cannot ever divide by zero. This is an undefined function in mathematics and algebra. We need another method to figure out how to take this limit. We are allowed to manipulate the function algebraically as long as we do not break any math rules. Notice that the denominator is factorable

$$\lim_{x\rightarrow 1}\frac{x-1}{x^2-1} = \lim_{x\rightarrow 1} \frac{x-1}{(x-1)(x+1)}$$

Now we can see that (x-1) is found in the numerator and denominator. We can simplify the expression into

$$\lim_{x\rightarrow 1}\frac{1}{x+1}$$

And now if we do the direct injection of x=1 we get

$$\lim_{x\rightarrow 1} \frac{1}{x+1} = \frac{1}{2}$$

And we have discovered the limit!

## Conclusion

The limit is the behavior of a function as the variable approaches a specific number. Limits can be found in numerous different ways, this post has shown you two specific methods to discover limits:

1. Table of values: Take values getting really close to the value you are searching for and measure the behavior of f(x).
2. Direct Injection: Try plugging in the value you are searching for directly into f(x), and if it fails, try manipulating the equation using standard algebra techniques.

Check back soon for more information on Limits and Calculus in general!

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