WEBVTT
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Properties of Limits
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In this video, we will learn how to use the properties of limits such as the limits of sums, differences, products, and quotients of functions and the limits of certain composite functions.
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Weβll be looking at various examples of how we can use these properties.
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Letβ²s start by defining some properties.
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Suppose that π of π₯ and π of π₯ are functions and π is some value such that the limit as π₯ tends to π of π of π₯ and the limit as π₯ tends to π of π of π₯ both exist.
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Then, we have the property for limits of sums of functions, which tells us that the limit as π₯ tends to π of π of π₯ plus π of π₯ is equal to the limit as π₯ tends to π of π of π₯ plus the limit as π₯ tends to π of π of π₯.
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We also have a property for the limits of differences of functions.
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And this tells us that the limit as π₯ tends to π of π of π₯ minus π of π₯ is equal to the limit as π₯ tends to π of π of π₯ minus the limit as π₯ tends to π of π of π₯.
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And we can note that these two properties can be used in combination with one another.
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Letβ²s now look at an example of how these properties can be used.
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Given that the limit as π₯ tends to two of π of π₯ is equal to three, the limit as π₯ tends to two of π of π₯ is equal to negative seven, and the limit as π₯ tends to two of β of π₯ is equal to negative one, find the limit as π₯ tends to two of π of π₯ plus π of π₯ minus β of π₯.
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In order to find this limit, we can start by breaking it down using the properties of limits.
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We have the property for the limits of sums of functions which tells us that the limit as π₯ tends to π of π of π₯ plus π of π₯ is equal to the limit as π₯ tends to π of π of π₯ plus the limit as π₯ tends to π of π of π₯.
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We also have the property for the limits of differences of functions.
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And this tells us that the limit as π₯ tends to π of π of π₯ minus π of π₯ is equal to the limit as π₯ tends to π of π of π₯ minus the limit as π₯ tends to π of π of π₯.
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Now, we can apply these two properties to the limit weβre trying to find.
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We can start by using the rule for limits of sums of functions.
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In our case, π is equal to two.
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And we can split the inside of our limit up so that we are adding two functions, these functions being π of π₯ and π of π₯ minus β of π₯.
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We obtain that our limit is equal to the limit as π₯ tends to two of π of π₯ plus the limit as π₯ tends to two of π of π₯ minus β of π₯.
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Next, we can use the rule for the limits of differences of functions.
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Again, π is equal to two.
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And inside our limit, we have a difference of two functions that being π of π₯ and β of π₯.
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Applying the rule, we obtain that our limit is equal to the limit as π₯ tends to two of π of π₯ plus the limit as π₯ tends to two of π of π₯ minus the limit as π₯ tends to two of β of π₯.
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Now we can spot that we know the value of each of these three limits since they have been given to us in the question.
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So substituting in three, negative seven, and negative one, we obtain that our limit is equal to three plus negative seven minus negative one.
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Simplifying this, we obtain a solution that the limit as π₯ tends to two of π of π₯ plus π of π₯ minus β of π₯ is equal to negative three.
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Now letβ²s cover some more properties of limits.
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Again, we have functions π of π₯ and π of π₯ with some constant values π and π such that the limit as π₯ tends to π of π of π₯ and the limit as π₯ tends to π of π of π₯ both exist.
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This time, we have an extra constant π.
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And weβ²ll see why this is here in our first property.
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This first property is about multiplicative constants within a limit.
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It tells us that the limit as π₯ tends to π of π multiplied by π of π₯ is equal to π multiplied by the limit as π₯ tends to π of π of π₯.
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Essentially, what this tells us is that if we have a constant factor within our limit, we can simply factor it outside of the limit.
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The next property is the limit of product of functions.
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It tells us that the limit as π₯ tends to π of π of π₯ multiplied by π of π₯ is equal to the limit as π₯ tends to π of π of π₯ multiplied by the limit as π₯ tends to π of π of π₯.
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Our third property here is for the limits of quotients of functions.
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It tells us that the limit as π₯ tends to π of π of π₯ over π of π₯ is equal to the limit as π₯ tends to π of π of π₯ over the limit as π₯ tends to π of π of π₯.
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And itβ²s again important to note that each of these rules can be used in combination with one another, including the two rules we covered previously.
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Letβ²s now look at some examples of how these rules can be used.
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Assume that the limit as π₯ tends to three of π of π₯ is equal to five, the limit as π₯ tends to three of π of π₯ is equal to eight, and the limit as π₯ tends to three of β of π₯ is equal to nine.
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Find the limit as π₯ tends to three of π of π₯ multiplied by π of π₯ minus β of π₯.
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We can start by breaking down the limit given in the question using the properties of limits.
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Firstly, we can use the rule for the limits of differences of functions.
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This tells us that the limit as π₯ tends to π of π of π₯ minus π of π₯ is equal to the limit as π₯ tends to π of π of π₯ minus the limit as π₯ tends to π of π of π₯.
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Looking at our limit, we can see that our value of π is three.
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And we can see that we have a difference of functions within our limit.
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We have π of π₯ multiplied by π of π₯ minus β of π₯.
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So we can apply our rule, giving us that our limit is equal to the limit as π₯ tends to three of π of π₯ times π of π₯ minus the limit as π₯ tends to three of β of π₯.
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In order to split down this limit further, weβll need to use another limit property.
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And this is the property of limits of products of functions, which tells us that the limit as π₯ tends to some constant π of a product of functions β so π of π₯ times π of π₯ β is equal to the limit as π₯ tends to π of π of π₯ times the limit as π₯ tends to π of π of π₯.
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Again, our value of π is three.
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And we can see that inside our limit, we have a product of functions.
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So thatβs π of π₯ times π of π₯.
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Applying this property, we find our limit is equal to the limit as π₯ tends to three of π of π₯ multiplied by the limit as π₯ tends to three of π of π₯ minus the limit as π₯ tends to three of β of π₯.
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And we can spot that weβ²ve been given each of these three limits within the question.
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We have that the limit as π₯ tends to three of π of π₯ is equal to five, the limit as π₯ tends to three of π of π₯ is equal to eight, and the limit as π₯ tends to three of β of π₯ is equal to nine.
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So we substitute these values in for our limit, giving us that our limit is equal to five timesed by eight minus nine.
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This can be simplified to obtain a solution that the limit as π₯ tends to three of π of π₯ multiplied by π of π₯ minus β of π₯ is equal to 31.
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In the next example, we will see how the property for quotients of functions can be used.
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Given that the limit as π₯ tends to negative two of π of π₯ over three π₯ squared is equal to negative three, determine the limit as π₯ tends to negative two of π of π₯.
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In this question, weβ²ve been given the limit as π₯ tends to negative two of π of π₯ over three π₯ squared.
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We can break this limit down using the properties of limits.
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We have the property for limits of quotients of functions, which tells us that the limit as π₯ tends to π of π of π₯ over π of π₯ is equal to the limit as π₯ tends to π of π of π₯ over the limit as π₯ tends to π of π of π₯.
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For our limit, weβre taking the limit as π₯ tends to negative two.
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Therefore, π is equal to negative two.
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And we have a quotient of functions.
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In the numerator, we have π of π₯.
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And in the denominator, we have three π₯ squared.
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Applying this rule for limits of quotients of functions, we obtain that our limit is equal to the limit as π₯ tends to negative two of π of π₯ over the limit as π₯ tends to negative two of three π₯ squared.
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Letβ²s now consider the limit in the denominator of the fraction.
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Thatβ²s the limit as π₯ tends to negative two of three π₯ squared.
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We can apply direct substitution to this limit, giving us that the limit as π₯ tends to negative two of three π₯ squared is equal to three times negative two squared.
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Negative two squared is equal to four.
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We then simplify to obtain that this limit is equal to 12.
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We can substitute this value of 12 back in to the denominator of our fraction, giving us that the limit as π₯ tends to negative two of π of π₯ over three π₯ squared is equal to the limit as π₯ tends to negative two of π of π₯ all over 12.
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However, weβ²ve been given in the question that the limit as π₯ tends to negative two of π of π₯ over three π₯ squared is equal to negative three.
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And since this is on the left-hand side of our equation, we can set our equation equal to negative three.
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So we now have that the limit as π₯ tends to negative two of π of π₯ over 12 is equal to negative three.
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We simply multiply both sides of the equation by 12.
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Here, we reach our solution which is that the limit as π₯ tends to negative two of π of π₯ is equal to negative 36.
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There are a couple more limit properties weβll be covering in this video and these are as follows.
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Given some function π of π₯ with some value π and an integer π such that the limit as π₯ tends to π of π of π₯ exists.
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We have the property for the limits of powers of functions, which tells us that the limit as π₯ tends to π of π of π₯ to the power of π is equal to the limit as π₯ tends to π of π of π₯ all to the power of π.
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So essentially, if we have a power of a function within a limit, we can take the power outside of the limit and simply raise the whole limit to that power.
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Letβ²s quickly note that our integer value of π here can be both positive or negative.
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And this rule will still work.
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We can see how this limit property can be derived from the limits of products of functions and the limits of quotients of functions properties.
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Since if we repeated either of those properties over and over again π times with just one function π of π₯, then we would obtain this property.
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Our final limit property here is the property for the limits of roots of functions.
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It tells us that the limit as π₯ tends to π of the πth root of π of π₯ is equal to the πth root of the limit as π₯ tends to π of π of π₯.
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So essentially, if we have an πth root of a function within our limit, then we can take the πth root outside of the limit and instead take the πth root of limit of the function.
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These two properties can again be combined with one another and with any of the previous properties.
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Letβ²s look at some examples of how they can be used.
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Assume that the limit as π₯ tends to six of π of π₯ is equal to three and the limit as π₯ tends to six of π of π₯ is equal to eight.
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Find the limit as π₯ tends to six of the square root of π of π₯ minus π of π₯.
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We need to find the limit as π₯ tends to six of the square root of π of π₯ minus π of π₯.
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We can break this limit down using the properties of limits.
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We have the property for the limits of roots of functions.
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It tells us that the limit as π₯ tends to some constant π of the πth root of some function π of π₯ is equal to the πth root of the limit as π₯ tends to π of π of π₯.
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The limit weβre trying to find is the limit as π₯ tends to six of the square root of π of π₯ minus π of π₯.
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So we have the limit of a square root of a function.
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We can therefore apply our rule for limits of roots of functions.
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It tells us that our limit is equal to the square root of the limit as π₯ tends to six of π of π₯ minus π of π₯.
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Now, we can see that we have the limit of a difference of functions since our limit is of π of π₯ minus π of π₯.
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We can apply the rule for the limit of differences of functions, which tells us that the limit as π₯ tends to some constant π of a difference of functions β so thatβs π of π₯ minus π of π₯ β is equal to the limit as π₯ tends to π of π of π₯ minus the limit as π₯ tends to π of π of π₯.
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We can apply this rule to our limit within the square root, giving us that our limit is equal to the square root of the limit as π₯ tends to six of π of π₯ minus the limit as π₯ tends to six of π of π₯.
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Now, we can spot that the limits within our square root have been given to us in the question.
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We have that the limit as π₯ tends to six of π of π₯ is equal to three and the limit as π₯ tends to six of π of π₯ is equal to eight, giving us that our limit is equal to the square root of eight minus three.
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Simplifying this, we obtain our solution which is that the limit as π₯ tends to six of the square root of π of π₯ minus π of π₯ is equal to the square root of five.
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Next, weβ²ll move onto our final example, where we will see how the properties of limits can be used even when the function is defined graphically.
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Consider the graph of π of π₯.
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Find the limit as π₯ tends to one of π₯ multiplied by π of π₯ all squared.
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Here, weβve been asked to find the limit as π₯ tends to one of π₯ multiplied by π of π₯ all squared.
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We can see that we have a power of a function.
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And so, we can use our rule for the limits of powers of functions.
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It tells us that the limit as π₯ tends to π of π of π₯ to the power of π is equal to the limit as π₯ tends to π of π of π₯ all to the power of π.
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In the case of our limit, the value of π is one and the power which we are raising our function to is two.
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So π is equal to two.
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Now, we can apply this rule.
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It tells us that our limit is equal to the limit as π₯ tends to one of π₯ multiplied by π of π₯ all squared.
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Next, we can in fact simplify the limit within our square further.
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We have the rule for the limits of product of functions.
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It tells us that the limit as π₯ tends to π of π of π₯ multiplied by π of π₯ is equal to the limit as π₯ tends to π of π of π₯ multiplied by the limit as π₯ tends to π of π of π₯.
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Now our product of functions is π₯ multiplied by π of π₯.
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So when we apply this rule to our limit, we obtain the limit as π₯ tends to one of π₯ multiplied by the limit as π₯ tends to one of π of π₯ all squared.
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Now, letβ²s consider the limit as π₯ tends to one of π₯.
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We can apply direct substitution to this limit.
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And we obtain that it is equal to one.
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So we can substitute this back in for our limit, giving us that the limit as π₯ tends to one of π₯ timesed by π of π₯ squared is equal to the limit as π₯ tends to one of π of π₯ all squared.
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Now all we need to do is find the limit as π₯ tends to one of π of π₯.
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In order to do this, we need to use our graph.
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We need to find the value of π of π₯ when π₯ is equal to one.
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We see that when π₯ is equal to one, π of π₯ is equal to three.
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And the graph of π of π₯ near to the π₯-value of one is a straight line.
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Therefore, the right limit of π of π₯ and the left limit of π of π₯ will both agree that this limit is equal to three.
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So we can substitute in three for the limit as π₯ tends to one of π of π₯.
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So we find that our limit is equal to three squared.
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And we can square the three to obtain our solution that the limit as π₯ tends to one of π₯ multiplied by π of π₯ squared is equal to nine.
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Now we have covered a variety of examples letβ²s recap some key points of the video.
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Key Points
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For functions π of π₯ and π of π₯ with values π and π and integer π such that the limit as π₯ tends to π of π of π₯ and the limit as π₯ tends to π of π of π₯ exist: the limit as π₯ tends to π of π of π₯ plus π of π₯ is equal to the limit as π₯ tends to π of π of π₯ plus the limit as π₯ tends to π of π of π₯.
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The limit as π₯ tends to π of π of π₯ minus π of π₯ is equal to the limit as π₯ tends to π of π of π₯ minus the limit as π₯ tends to π of π of π₯.
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The limit as π₯ tends to π of π multiplied by π of π₯ is equal to π multiplied by the limit as π₯ tends to π of π of π₯.
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The limit as π₯ tends to π of π of π₯ multiplied by π of π₯ is equal to the limit as π₯ tends to π of π of π₯ multiplied by the limit as π₯ tends to π of π of π₯.
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The limit as π₯ tends to π of π of π₯ over π of π₯ is equal to the limit as π₯ tends to π of π of π₯ over the limit as π₯ tends to π of π of π₯.
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The limit as π₯ tends to π of π of π₯ to the power of π is equal to the limit as π₯ tends to π of π of π₯ all to the power of π.
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The limit as π₯ tends to π of the πth root of π of π₯ is equal to the πth root of the limit as π₯ tends to π of π of π₯.
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And all of these limit properties can be used in combination with one another.