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(a) What is wrong with the following equation?$$ \frac {x^2 + x - 6}{x - 2} = x + 3 $$

(b) In view of part (a), explain why the equation $$ \lim_{x \to 2}\frac {x^2 + x - 6}{x - 2} = \lim_{x \to 2}(x + 3) $$is correct.

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04:35

Daniel Jaimes

03:43

Stark Ledbetter

Calculus 1 / AB

Chapter 2

Limits and Derivatives

Section 3

Calculating Limits Using the Limit Laws

Limits

Derivatives

Missouri State University

University of Michigan - Ann Arbor

Boston College

Lectures

04:40

In mathematics, the limit of a function is the value that the function gets very close to as the input approaches some value. Thus, it is referred to as the function value or output value.

In mathematics, a derivative is a measure of how a function changes as its input changes. Loosely speaking, a derivative can be thought of as how much one quantity is changing in response to changes in some other quantity; for example, the derivative of the position of a moving object with respect to time is the object's velocity. The concept of a derivative developed as a way to measure the steepness of a curve; the concept was ultimately generalized and now "derivative" is often used to refer to the relationship between two variables, independent and dependent, and to various related notions, such as the differential.

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we had the expression X squared plus x minus six over x minus two. We can factor the numerator. The numerator would factor into X plus straight Times X -2. And of course in the denominator, we have X -2. Now we'd like to cancel out these X -2s. Uh but we can do so under one assumption That X is not equal to two. If X was equal to two, then the X -2 that you did have in the denominator before you try to cancel it out would have been zero and you can't have zero in the denominator, you cannot divide by zero. So X is not allowed to equal to As long as X is not allowed to equal to, you can cancel the X -2s and get x plus straight. So it's not as simple as saying that this expression equals X plus three. This expression does equal X-plus three. As long as you add the condition X is not allowed to equal to because remember effects was to the original denominator would have been zero. Can't divide by zero. No. If we want to take the limit of this function as X approaches to Now, if we want to take the limit of this original function as X approaches to uh then we can uh find that limit by taking the limit as X approaches to of the X plus straight. The reason we're allowed to do that is when we find a limit of this function as X approaches to. The limit is only concerned with X approaching to um X does not become two. X is not does not get to equal to when we're finding limits, we're only interested interested in what is happening as X approaches to. So as long as uh since finding the limit as X approaches to, we're not going to be having X equal to two, then we're allowed to have this function written as X plus three because this function was allowed to equal X plus three as long as X was not equal to two. When we're finding the limit as X approaches to, we're not going to have X equaling to only X approaching two. So the limit of this function as X approaches to is simply equal to the limit of this simpler function as X approaches to. And of course this is a nice continuous function. So we can find this limit as X approaches to simply by substituting two and for X. So the limit of X plus three as X approaches to is two plus three, Of course, that's five.

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