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In this video, we will learn how to use the direct substitution method to evaluate limits.
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There are certain conditions which need to be met in order to use direct substitution.
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Weβll be covering these conditions and looking at some examples.
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Letβs start by recalling the definition of a limit.
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We say that, for some function π of π₯, which is defined near π, the limit of π of π₯ as π₯ approaches π is πΏ.
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What this means is that the closer π₯ gets to π, the closer the value of π of π₯ gets to πΏ.
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Formally, we write this like this.
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The limit as π₯ goes to π of π of π₯ is equal to πΏ.
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In order to find a limit using direct substitution, we simply substitute π₯ equals π into π of π₯.
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And so the result we get for direct substitution is that the limit as π₯ goes to π of π of π₯ is equal to π of π.
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This is what weβll be using in order to find limits using direct substitution.
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However, there are only certain cases when weβre allowed to use direct substitution in order to find a limit.
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The functions which weβre taking the limit of β so thatβs π of π₯ β must satisfy at least one of the following conditions.
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The first condition which will allow us to use direct substitution is that if π of π₯ is a polynomial, that is, if π of π₯ is of the form given here, where π naught to π π are all constants.
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Now remember that this also includes constant functions.
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The second condition which could enable us to use direct substitution is if π of π₯ is a rational function.
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This means that π of π₯ is equal to some π of π₯ over π of π₯, where π of π₯ and π of π₯ are both polynomials.
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And we also require that π of π is nonzero, since if π of π is equal to zero, then the denominator of π of π will be equal to zero.
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And so π of π will be undefined.
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The third condition is that π of π₯ is either a trigonometric, exponential, or logarithmic function.
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For the fourth condition, we have that π of π₯ is a power function.
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So this means that π of π₯ is equal to π₯ to the power of π, where π is a real number.
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Letβs note that this also includes negative and fractional powers, such as π₯ to the power of negative one-half, which is also equal to one over the square root of π₯.
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For the fifth condition, we have that if π of π₯ is a sum, difference, product, or quotient of functions for which direct substitution works.
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So that could be a combination of any of the types of functions weβve already covered in these conditions.
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The final condition is if π of π₯ is a composition of functions for which direct substitution works.
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What this means is that if π of π₯ is equal to π of β of π₯, where β of π₯ allows substitution at π and π of π₯ allows substitution at β of π.
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Now we have covered all the conditions under which we can use direct substitution in order to solve a limit.
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Weβre now ready to look at an example.
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Determine the limit as π₯ goes to negative five of negative nine π₯ squared minus six π₯ minus nine.
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Here weβre asked to find the limit of the function negative nine π₯ squared minus six π₯ minus nine.
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And so we can write this down as π of π₯.
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We can see that π of π₯ is simply a polynomial.
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Therefore, weβre able to use direct substitution in order to find this limit.
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Direct substitution tells us that the limit of π of π₯ as π₯ goes to π is equal to π of π.
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Applying this to our question, we can say that the limit of π of π₯ as π₯ goes to negative five is equal to π of negative five.
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In order to reach our solution, we simply substitute negative five into π of π₯.
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This gives us negative nine multiplied by negative five squared minus six multiplied by negative five minus nine.
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You may find it helpful to put brackets around the negative numbers so that when we multiply the negative numbers, we do not forget about the negative sign.
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Firstly, letβs expand the negative five squared.
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If we remembered that a negative number timesed by a negative number gives a positive number, then we will get that negative five squared is equal to 25.
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So we can write negative nine multiplied by 25.
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Next, we can multiply the negative six by the negative five, to get positive 30.
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And then we simply have to subtract the nine on the end.
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Multiplying the negative nine by the 25 gives us negative 225.
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And we subtract the nine from the 30 to give us plus 21.
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Adding 21 to negative 225 gives us a solution of negative 204.
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In this example, we saw how to apply direct substitution when finding the limit of a polynomial function.
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Next, weβll be considering some functions and trying to work out whether or not they satisfy the conditions to use direct substitution.
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Which of the following functions satisfies the conditions for direct substitution of the limit, the limit of π of π₯ as π₯ goes to zero?
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A) π of π₯ is equal to π₯ squared plus five π₯ over π₯ squared minus two π₯.
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B) π of π₯ is equal to π₯ squared minus five π₯ plus six over π₯ minus two sin π₯.
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C) π of π₯ is equal to π₯ if π₯ is greater than three and π₯ minus three if π₯ is less than or equal to three.
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D) π of π₯ is equal to π₯ plus one over π₯.
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And E) π of π₯ is equal to two π₯ if π₯ is greater than zero and two π₯ minus one if π₯ is less than or equal to zero.
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In order to find which of these functions we can use direct substitution in order to find the limit, we must test the conditions for which direct substitution works on each of the functions.
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For function A, we have π of π₯ is equal to π₯ squared plus five π₯ over π₯ squared minus two π₯.
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This is a rational function.
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And if we write π of π₯ as π of π₯ over π of π₯, then we obtain that π of π₯ is equal to π₯ squared plus five π₯ and π of π₯ is equal to π₯ squared minus two π₯.
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Since both π of π₯ and π of π₯ are polynomials, the only condition which we need to check is that the denominator of the fraction is nonzero at π₯ equals zero.
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The denominator of our function is π of π₯.
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So letβs work out what π of zero is.
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We simply substitute zero into π₯ squared minus two π₯, which gives us zero squared minus two times zero.
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And zero squared and two times zero are both zero.
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So therefore, π of zero is equal to zero.
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This means that the denominator of our function is equal to zero at π₯ equals zero.
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And so weβre not able to use direct substitution in order to find this limit.
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Moving on to function B, we have π of π₯ is equal to π₯ squared minus five π₯ plus six over π₯ minus two sin π₯.
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And this here is a quotient of two functions.
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So we can write π of π₯ as π of π₯ over π of π₯, where π of π₯ is equal to π₯ squared minus five π₯ plus six and π of π₯ is equal to π₯ minus two sin π₯.
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In order to use direct substitution to find the limit of this function, we again need to check what the denominator equals at π₯ equals zero.
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We find π of zero is equal to zero minus two multiplied by sin of zero.
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And this is also equal to negative two sin of zero.
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However, sin of zero is equal to zero.
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And this gives us that π of zero must also be equal to zero.
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So similarly to function A, we have found that the denominator of function B is also zero at π₯ equals zero.
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And therefore, again, we cannot use direct substitution in order to find the limit of this function.
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Moving on to function C, we have that π of π₯ is equal to π₯ if π₯ is greater than three and π₯ minus three if π₯ is less than or equal to three.
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In order to find whether we can use direct substitution to find the limit of this function, we first need to consider whether the limit actually exists for this function.
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In order for the limit to exist, we require that the one-sided limits on either side of zero are equal.
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So this means that the limit as π₯ approaches zero from above of π of π₯ must be equal to the limit as π₯ approaches zero from below of π of π₯.
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As π₯ approaches zero from above, we have that π of π₯ will be equal to π₯ minus three.
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And this is because when π₯ is just larger than zero, π₯ will still be less than or equal to three.
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And so, therefore, our function π of π₯ will be equal to π₯ minus three.
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And this is just a polynomial function.
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And so we can use direct substitution in order to find the limit from above.
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And so we find the limit as π₯ approaches zero from above of π of π₯ is equal to π of zero, which is also equal to zero minus three, or just negative three.
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Next, if we consider the limit as π₯ approaches zero from below, we know that π of π₯ is again equal to π₯ minus three, since zero is less than or equal to three, meaning that our π₯-value is less than or equal to three.
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So therefore, π of π₯ must be equal to π₯ minus three.
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Again, this is a polynomial.
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So we will use direct substitution in order to find the limit as π₯ approaches zero from below, giving us that the limit is equal to π of zero, which is again equal to zero minus three, or negative three.
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And so we have found that the limit as π₯ approaches zero from above and the limit as π₯ approaches zero from below are equal.
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And therefore, it satisfies our condition for the limit to exist.
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We can add on to the end of our condition that these two one-sided limits will also be equal to the limit as π₯ approaches zero of π of π₯.
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Now we know that the limit exists, we just need to check that we can use direct substitution to solve it.
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At π₯ equals zero, π of π₯ is equal to π₯ minus three, which is a polynomial function, meaning that we can use direct substitution in order to find the limit of this function.
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Function D is π of π₯ is equal to π₯ plus one over π₯.
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This is again a rational function.
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And we can again say that itβs equal to π of π₯ over π of π₯, where π of π₯ is equal to π₯ plus one and π of π₯ is just equal to π₯.
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We substitute π₯ equals zero into the denominator of the fraction, which is π of π₯.
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We find that π of zero is equal to zero.
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Therefore, the denominator of π of π₯ is equal to zero at π₯ equals zero.
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Direct substitution can therefore not be used in order to find this limit.
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For function E, we have π of π₯ is equal to two π₯ if π₯ is greater than zero and two π₯ minus one if π₯ is less than or equal to zero.
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For this piecewise function, we again need to consider the one-sided limits to the left and right of zero in order to check whether the limit itself exists.
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Remembering that, in order for the limit to exist, the limit as π₯ approaches zero from above must be equal to the limit as π₯ approaches zero from below of π of π₯.
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When π₯ is approaching zero from above β so this means π₯ is just a bit larger than zero β we have that π of π₯ is equal to two π₯, since π₯ is just larger than zero.
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π of π₯ is equal to two π₯ is just a polynomial.
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And so we can find the limit as π₯ approaches zero from above using direct substitution.
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We find that itβs equal to π of zero, which is equal to two times zero, or just zero.
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Next, weβll consider the limit as π₯ approaches zero from below.
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When π₯ is smaller than or equal to zero, we have that π of π₯ is equal to two π₯ minus one, which is again just a polynomial.
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We can therefore find the limit using direct substitution.
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This gives us that the limit as π₯ approaches zero from below of π of π₯ is equal to π of zero, or two times zero minus one.
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And since two times zero is just zero, we get that this is equal to negative one.
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Now if we compare these one-sided limits, we can see that theyβre not equal to one another.
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This tells us that the limit does not exist here.
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And so we cannot use direct substitution in order to find this limit.
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We find that the solution to this question is C.
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Given π of π₯ is equal to the modulus of π₯ plus 11 minus the modulus of π₯ minus 18, find the limit as π₯ goes to four of π of π₯.
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Now our list of conditions under which direct substitution works does not include the modulus function.
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However, we can think of the modulus function in another way.
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We can write the modulus of π₯ as the square root of π₯ squared.
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Since finding the modulus of a number is simply taking the absolute value of that number and taking the square and then the root of a number will also give us the absolute value of that number.
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The square root of π₯ squared can also be written as π₯ squared to the power of one-half.
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π₯ squared and π₯ to the power of a half are both power functions.
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We know that we can apply direct substitution to power functions.
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And π₯ squared to the power of a half is simply a composition of two power functions.
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Therefore, we can also apply direct substitution to this.
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From this, we can see that we can apply direct substitution to the function mod π₯.
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Now weβre ready to consider the function π of π₯.
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We will consider the modulus of π₯ plus 11 and the modulus of π₯ minus 18.
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Here weβre simply taking the moduluses of two polynomial functions, which are π₯ plus 11 and π₯ minus 18.
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This is simply a composition of a polynomial function and a modulus function.
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And since we know we can apply direct substitution to both polynomial functions and modulus functions, we know that we can also apply direct substitution to these composite functions.
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Now π of π₯ is simply the difference of these two composite functions.
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Therefore, we can also apply direct substitution to π of π₯.
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So the limit as π₯ approaches four of π of π₯ is equal to π of four, which is also equal to the modulus of four plus 11 minus the modulus of four minus 18, or mod 15 minus mod of negative 14, which is also equal to 15 minus 14.
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From this, we find that the solution to this question is simply one.
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Now we have seen a variety of limits which can be found using direct substitution and some which cannot.
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Let us recap some of the key points of this video.
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We have that the formula for using a direct substitution is that the limit of π of π₯ as π₯ approaches π is equal to π of π.
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In order to use direct substitution, the limit must exist and the function must be defined at the point at which weβre taking the limit.
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And finally, the function weβre taking the limit of must be a polynomial, rational, trigonometric, exponential, logarithmic, or power function or a sum, difference, product, quotient, or composition of any of these types of functions.