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Continuity of Functions
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In this lesson, weβll learn how to check the continuity of a function and to determine the interval on which a function is continuous.
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Now, you may be familiar with finding the continuity of a function at a point and, indeed, the different types of discontinuity that we come across.
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We can test whether a function is continuous at a point using the following formal definition.
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A function π of π₯ is continuous at the point where π₯ is equal to π if the limit, as π₯ approaches π of π of π₯, is equal to the function evaluated at π₯ equals π.
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Now, the so-much-implied requirement to this condition is that both of these things must exist.
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In order for the normal limit to exist, the left and the right limit must also exist as π₯ approaches π and be equal to some value πΏ.
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Alongside this, π of π must be defined, and it must be equal to the left, the right, and the normal limit for continuity.
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Here, weβve said this value is πΏ.
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Now, letβs think about continuity over an interval.
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The colloquial definition of this might be, if youβre able to draw the graph of the function over the interval without lifting your pen.
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A more rigorous way to think about this would be to say, the function π of π₯ is continuous over an interval if the requirement for continuity at a point holds for all values of π₯ within the interval.
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Now, this may seem obvious, but one way to check for continuity over an interval is to ensure that there are no discontinuities in said interval.
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We can look at a couple of example graphs to get a visual understanding of this.
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Determine whether the following statement is true or false.
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The function represented by the graph is a continuous function.
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For this question, weβve been given a function π of π₯, which is defined for all values of π₯ over the real numbers, denoted by these arrows.
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Now, we can almost immediately see that our function π of π₯ is discontinuous, since at a value of π₯ equals three, we have a gap in our graph.
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In fact, we may recognise this as a jump discontinuity, with π of π₯ being undefined at the point three, two, denoted by the hollow dot, and defined at the point three, one, denoted by the filled dot.
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If we were to look at the left- and right-sided limits, as π₯ approaches three, we would find that although both exist, their values disagree.
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And this would mean that the normal limit does not exist.
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Here, we recall our definition for continuity at a point, which says that the limit, as π₯ approaches π, of said function must be equal to the value of the function evaluated where π₯ is equal to π.
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Now, in our case, the limit, as π₯ approaches three of π of π₯, is not equal to π of three, since the limit does not exist.
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We have, therefore, proved that the function π of π₯ has a discontinuity at π₯ equals three.
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Our answer to the question is, therefore, false.
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The function represented by the graph is not a continuous function.
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As a final point, we may note that our function can have a discontinuity even though it is defined over all real numbers.
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Letβs now look at another graphical example of determining whether a function is continuous or not.
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Determine whether the function represented by the graph is continuous or discontinuous.
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For this question, weβve been given a function π of π₯, which is defined when π₯ is greater than or equal to zero or less than or equal to three.
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The interesting points of this function occur when π₯ is equal to one and when π₯ is equal to two.
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Here, we see a sharp change in gradient.
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And we may recognise that this means our function would not be differentiable at these points.
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For the purpose of continuity, however, this is not necessarily cause for concern.
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Taking the point where π₯ equals one as an example, we see that both the left- and the right-sided limits approach the same value.
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And this will be where π of π₯ is equal to one.
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From this, it follows that the normal limit, as π₯ approaches one, will take the same value.
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And itβs also clear to see that π of one is also equal to one.
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In fact, these two things together are the condition for continuity.
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Since we have that the normal limit, as π₯ approaches one of π of π₯, is equal to π of one.
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The same logic would follow for the point where π₯ is equal to two, and, in fact, for all other points along our functionβs domain.
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This puts us in a position to answer our question.
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We conclude that π of π₯ is a continuous function.
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Now, the two previous questions gave us graphical examples of functions which were continuous and functions which were not.
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In our example, we saw a function with a jump discontinuity.
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However, functions with any other discontinuity, such as removable or infinite, would also not be classed as continuous functions.
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Conversely, there are many types of functions which are continuous.
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And weβre gonna look at a few examples algebraically of evaluating these.
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The following types of functions are continuous across their entire domain, polynomial functions, rational functions, trigonometric functions, and exponential functions.
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An important distinction here is weβre saying that these functions are continuous over their entire domain and not for all values of π₯ over the real numbers.
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Weβll come back to this point later.
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But before we do, another important fact.
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Sums, differences, products, quotients, and compositions of continuous functions are also continuous for all points where π₯ is properly defined.
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Again, these are within the domains of our newly created functions.
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Now, the proof of continuity for all of these types of functions is outside the scope of this video, but we can go through one such example using a polynomial function and the following question.
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What can be said about the continuity of the function π₯ cubed plus five π₯ squared minus two π₯ plus two?
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Given the general rule that a polynomial function is continuous across its domain and the domain of our function is all of the real numbers, we can almost immediately give the following answer.
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The function is continuous on β, the real numbers, because it is a polynomial.
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This is, in fact, the quick answer to our question.
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But instead of just stopping here, letβs explore a striped-back proof of why this is the case.
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The building block that weβll start with is the limit, as π₯ approaches π, of just some constant π.
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Now, clearly, the value of π₯ doesnβt matter to our constant π.
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And so, the answer to this limit is just π.
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Here, weβll say that π is a real number.
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Next, we move on to the case of the limit, as π₯ approaches π, of the function which is just π₯.
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By taking a direct substitution approach of π₯ equals π, we see that, here, the answer to our limit is just π.
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And also we should here stipulate that π is also one of the real numbers.
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Next, we add a power and we take the limit as π₯ approaches π of π₯ raised to some power π.
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In this case, π is one of the natural numbers, including zero.
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So, zero, one, two, three, etcetera.
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With the same direct substitution argument, we find that this limit is equal to π to the power of π.
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What about if we added some constant π in front of our π₯ to the power of π?
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The constant multiple law allows us to move our constant π outside of our limit as follows.
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Now, we see the limit weβre looking for is the same as the previous line times π.
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And so, our answer is π times π to the power of π, again with the direct substitution approach.
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Okay, next, weβre gonna expand a bit by adding two new terms to our limit and differentiating between different values of π and π, where all values of π and π obey these same rules.
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Now, using the addition laws of limits, we can split our one limit into three individual limits as follows.
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Now, you may notice that these first two terms take the same form as the previous line of working that weβve just completed.
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We can, therefore, say that they are π one times π to the power of π one, and π two times π to the power of π two, respectively.
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Of course, our last term is just a constant, which we saw in our first example.
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The limit of this constant is, of course, just π three.
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We have now found that our limit is π one times π to the power of π one plus π two times π to the power of π two plus π three.
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Now, if we call the function that weβve created here π of π₯, we should be able to see that the limit that we have found is equal to our function evaluated at the point where π₯ is equal to π, in other words, π of π.
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Our final and crucial step is in recognising the form of the function π of π₯ that weβve created.
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Since, π is a real number and π is a natural number including zero, each of these terms gives us a real multiple of π₯ raised to any positive integer power or zero.
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By the limit rule of addition, we could add as many of these terms as we like to our function.
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Alongside this, weβve covered off the addition of a standard real constant π.
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Our function π of π₯, therefore, represents any polynomial that we care to construct.
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Finally, weβve proved that the limit, as π₯ approaches π, for our function π of π₯ is equal to π of π.
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Since π of π₯ is any polynomial and π is any real number, we have just proved that a polynomial function is continuous across the entire set of real numbers, since this is indeed the condition for continuity.
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Weβve now proved our continuity condition for polynomials.
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And this proof can be extended or altered to cover the other functions mentioned earlier.
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Now, when classifying that these functions were continuous, we were careful to make the distinction that theyβre continuous across their domain, and not for the entirety of the real numbers.
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As it happens, the domain of a polynomial is the real numbers, however, this is not necessarily the case for other functions such as rationals.
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Let us look at one such example to illustrate this.
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Find the set on which π of π₯ which is equal to π₯ minus 22 all over π₯ squared minus two π₯ minus 63 is continuous.
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For this question, we have a rational function in the form π of π₯ over π of π₯.
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Now, we know that a rational function is continuous over its domain.
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And hence, our question reduces to finding the domain of our function π of π₯.
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In essence, we want to find values of π₯ which would cause our function either to be undefined or to grow towards positive or negative infinity.
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Looking at the form of our function, we see that these troublesome points will occur when the denominator of our quotient, π of π₯, evaluates to zero.
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We can move forward with our question first by factorising π of π₯.
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With a bit of inspection, we see that this quadratic factorises to π₯ minus nine times π₯ plus seven since these two numbers have a sum of negative two and a product of negative 63.
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From the factor theorem, we can then see that when π₯ is nine or when π₯ is negative seven, π of π₯ will evaluate to zero.
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Putting this factorised version of π of π₯ back into our function, we can then conclude that π₯ equals nine and π₯ equals negative seven are not in the domain of our function.
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This is because, at these values, the denominator of our quotient would be zero.
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And, hence, π of π₯ would not give us a numerical evaluation.
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Since π of π₯ behaves at all other real values of π₯, we can say the following.
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The domain of our function π of π₯ is the real numbers minus the set of nine and negative seven.
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And so, π of π₯ is continuous over the real numbers minus the set of nine and negative seven.
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And here, we have answered our question.
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To expand upon our previous question, you may recall that when a common factor can be cancelled on the top- and bottom-half of a quotient, such as that forming a rational function, this would correspond to a removable singularity on our graph.
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In this case, since the function is undefined where π₯ equals π, clearly, we cannot meet the criteria for continuity.
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In cases where common factors cannot be cancelled on the top- and bottom-halves of our quotients, we would expect to see asymptotes where the values of π of π₯ would approach positive or negative infinity.
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At these asymptotes, even if the left- and the right-limits were to agree, such as in this case, although we may express that the limit, as π₯ approaches π of π of π₯, is equal to infinity, this is just a particular way of saying that the limit does not exist.
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And, in fact, the limit does not exist at any of these asymptotes.
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Now, since we have places where our limit does not exist, again, we cannot fulfill our criteria for continuity.
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And at these points, π of π₯ would be discontinuous.
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Moving on, some functions can be defined piecewise behaving differently over different intervals.
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The same rules of continuity apply within the intervals of our piecewise function, but we must be careful when examining the boundary between the intervals.
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In order for continuity to be maintained over the boundary, the end points of the two sections, or subfunctions, must join up.
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This is best illustrated with an example.
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Suppose that π of π₯ is equal to five sin of π₯ minus three over π₯ minus three if π₯ is less than three and five π₯ squared over nine if π₯ is greater than or equal to three.
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Find the set on which π is continuous.
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Here, we have a piecewise function defined over two intervals.
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The boundary of our two intervals occurs at the point where π₯ equals three.
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And hence, this is an important point.
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For our first subfunction, we have a trigonometric expression on the top-half of our quotient and a binomial on the bottom-half.
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Since we know that trigonometric functions and polynomials are continuous on their domains, and quotients of continuous functions are also continuous on their domains, we conclude that this subfunction is, therefore, also continuous on its domain.
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Now, here, we need to be slightly careful since, at values where π₯ equals three, this subfunction evaluates to zero over zero, which is an indeterminate form.
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Lucky for us, π of π₯ is only defined by this subfunction at values of π₯ which are strictly less than three, not where π₯ is equal to three.
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Instead, when π₯ is greater than or equal to three, π of π₯ is defined by five π₯ squared over nine.
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Again, here, itβs worth noting that this monomial is defined over and has a domain of all real numbers.
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This means that the domain of our function π of π₯ is all the real numbers.
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But we should be careful not to hastily jump to the conclusion that π of π₯ is also continuous over all the real numbers.
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Instead, we must still check the criteria for continuity at the boundary between our subfunctions, or when π₯ is equal to three.
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First, letβs find π of three by substituting in to five π₯ squared over nine.
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This value is easy to compute.
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And we get an answer of five.
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Next, we need to check that our normal limit exists and is also equal to five.
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If this is not the case, then weβll have a discontinuity at π₯ equals three.
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And hence, our function will not be continuous at this point.
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To move forward, we first recognise that, either side of π₯ equals three, our function is defined by two different subfunctions.
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To find our left limit, or when π₯ approaches from the negative direction, we use our first subfunction.
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Here, weβve already shown that a direct substitution of π₯ equals three leads us to an indeterminate form of zero over zero, so weβll need to use a different approach.
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Instead, we use the rule that the limit, as π₯ approaches zero of sin π₯ over π₯, is equal to one.
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Now, our expression isnβt in this form.
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And so, we perform some manipulations first by taking a factor of five outside of our limit using the constant multiple rule.
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Next, we perform a π’-substitution.
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By setting π’ as π₯ minus three, we get the following.
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Our limit becomes sin π’ over π’.
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However, we mustnβt forget to change the limit value itself.
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π’ plus three is equal to π₯.
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Hence, π’ plus three approaches three from the negative direction, or π’ approaches zero from the negative direction.
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Looking at our rule, we know that if the normal limit exists and is equal to one, then the left- and right-sided limits also exist and are also equal to one.
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We can now use our rule to evaluate that this limit is equal to one.
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Hence, the left-sided limit, as π₯ approaches three of π of π₯, is equal to five times one, which is, of course, just five.
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Now, our right-sided limit is far easier to evaluate.
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For this, since weβre approaching π₯ equals three from the right, we take the limit using our other subfunction.
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Simply, by direct substitution, we find that this limit is equal to five.
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Since the left- and right-sided limits both exist and are equal to the same value, we can, therefore, conclude that the limit, as π₯ approaches three of π of π₯, is also equal to five.
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And earlier, youβll recall that we found that π of three is also equal to five.
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Since the limit, as π₯ approaches three of π of π₯, is equal to π of three, we conclude that π of π₯ is continuous when π₯ is equal to three.
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If you were to think about this visually, this means that the two end points of our subfunctions will join up.
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Let us now recall that, earlier, we concluded that π of π₯ was continuous over all of the real numbers aside from three, which we had to check.
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And now that we have checked three, weβre in a position to say that our function π of π₯ is continuous on all the real numbers.
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And this is the answer to our question.
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To wrap things up, letβs go through a few key points.
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A function is continuous over an interval if it is continuous over all points within said interval.
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A function, which weβll call π of π₯, is continuous at a point, letβs say where π₯ equals π, if the limit, as π₯ approaches π of π of π₯, is equal to π of π.
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Here, we note the implication, firstly, that this limit exists and, secondly, that the function is defined when π₯ equals π.
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Polynomial, rational, trigonometric, and exponential functions are continuous over their domains.
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And here, we note that this is not necessarily all real numbers.
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Further to this, sums, differences, products, quotients, and compositions of continuous functions are also continuous at points where π₯ is properly defined.
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Discontinuities of a function can often be found by looking for values of x which result in a division by zero.
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And these may be removable or essential discontinuities.
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The boundary between the intervals of piecewise functions should be checked to make sure that the ends of the subfunctions join up.
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If the left- and the right-sided limits do not agree at these points, then the result will be a jump discontinuity.
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If the left- and the right-sided limits do agree, and theyβre both equal to the function evaluated at this point, then continuity will be maintained.