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Use properties of limits and algebraic methods to find the limit, if it exif x <-1lim f(x), where f(x)Jx? _ X - 1 ifx> -1...

Question

Use properties of limits and algebraic methods to find the limit, if it exif x <-1lim f(x), where f(x)Jx? _ X - 1 ifx> -1

Use properties of limits and algebraic methods to find the limit, if it ex if x <-1 lim f(x), where f(x) Jx? _ X - 1 ifx> -1



Answers

Use the properties of limits to help decide whether each limit
exists. If a limit exists, find its value.
$$\lim _{x \rightarrow 1} \frac{x^{2}-1}{x-1}$$

We are finding the limit of the following problem or the following function? Excuse me. So the limits X approaches one of F of X. Where And it's a piecewise function sort of So F of X is equal to X were placed to when X is not equal to one and F of X equals one when X is equal to one. So for these types of functions we want to evaluate the left hand side and the right hand side limit. So as X approaches one from the left of F of X. Um So think of numbers we are using numbers X, X values that are less than one close to one to make X approach one from the left. So if we plug one indirectly from that site, the limit will equal three. And as we approach um one from the right of ffx we are plugging in numbers that are not one but close to one but greater than one. So again we want to use the X square plus two branch because X is not one when you evaluate the limit, so the limit as you approach one from the right is going to also equal three since the Y values will be approaching three. So since the limit on both sides equal three, the limit of f f x s x approach one, purchase one is also equal to three, so therefore the one. And this problem does not matter because that is when X is equal to one. But since we are evaluating the limit, we want to use X values close to one.

Calculate the following limit at exposed to one. So if we do so we're gonna get um, x squared minus one is going to be once minus one is zero Over 0, 1 -1. This is indeterminant and that we can do two things. You can either do, um hospitals rule or we can simplify our fraction. So let's do the simplification. You're going to get the limit as X go Sue on of x minus one. Have X plus one. Getting difference of squares Over X -1 sex minus once castles, you get one Plus 1 or two. So the results of our dilemma.

In this question, we are trying to calculate the limits at X goes to one of the following fraction. Our fraction is x squared -1 over eggs -1. Not looking at the denominator as exposed to want. We're going to get 1 -1 or zero. Now getting zero had the denominator means that this is not allowed and we cannot solve the limit like this. So we need to rewrite our limits and try to simplify our fraction. The numerator is the difference of squares giving us x minus one and ex plus one Over X -1, X -1 from the top and the bottom cancel out living us with the limit at the X goes to juan of X plus one. nine X goes to want. The next plus one becomes 2. So that's our results.


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