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In this video, we will learn how to find the limits of trigonometric functions.
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We will be using some rules to help us.
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Letβs start by reminding ourselves what a limit is.
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If the limit of π of π₯ as π₯ approaches π exists, then we can say that itβs equal to some constant πΏ.
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And what this means is that π₯ approaches π.
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The function π of π₯ approaches πΏ.
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When finding the limits of trigonometric functions, there are some functions which can be found using direct substitution, for example, the limits of sin π₯ and cos π₯.
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Some functions require the use of trigonometric identities, such as the one shown here, in order to manipulate them into a form where we can use direct substitution.
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However, there are some cases where direct substitution does not work.
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One such case is the limit as π₯ approaches zero of sin π₯ over π₯.
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Since when we try to use direct substitution here, we obtain sin of zero over zero, which is also equal to zero over zero.
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And this is undefined.
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And so here weβve reached the first rule which weβll be using to help us find the limits of trigonometric functions.
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We have that the limit as π₯ approaches zero of sin π₯ over π₯ is equal to one.
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Now the proof of this rule is a bit beyond the scope of this video.
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However, if we think about this rule in a certain way, we can gain an intuitive insight into why it works.
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If we think about our small angle approximations, we know that when π₯ is very very small, sin of π₯ is roughly equal to π₯.
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In our limit, weβre taking the limit as π₯ approaches zero.
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So this means π₯ will be getting smaller and smaller.
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And therefore, it makes sense for us to use our small angle approximation here.
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We obtain that the limit as π₯ approaches zero of sin π₯ over π₯ is roughly equal to π₯ over π₯.
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Canceling the π₯s on the top and bottom of the fraction leaves us with one.
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Another way to think about this limit intuitively is to draw the graph of sin π₯ over π₯.
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We can see from plotting the graph that the line tends toward one at zero.
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We would obtain a similar result by considering a table of values for sin of π₯ over π₯.
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Now letβs consider an example using this identity.
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Evaluate the limit as π₯ approaches zero of sin of π₯ over sin of π₯ over two.
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First, we can try to solve this limit by using direct substitution.
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We substitute π₯ equals zero into our equation.
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However, this leaves us with zero over zero, which is undefined.
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We will need to find this limit by another means.
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Letβs try using the rule that the limit as π₯ approaches zero of sin π₯ over π₯ is equal to one.
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In order to obtain something of this form, we need to multiply both the numerator and denominator of our limit by π₯.
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In doing this, it enables us to write our limit as the limit as π₯ approaches zero of sin π₯ over π₯ multiplied by π₯ over sin of π₯ over two.
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Next, weβll be using the limit rule, which tells us that the limit of a product of functions is equal to the product of the limits of the functions.
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We obtain this.
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We notice that the limit on the left of the product is identical to the limit in our rule.
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And so we can say that this limit is just one.
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In order to evaluate the other limit, letβs rewrite our rule that, instead of writing π₯, weβll write π₯ over two.
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π₯ over two approaching zero is the same as π₯ approaching zero.
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And so we can write this in here.
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Next, we can multiply both the numerator and denominator by two.
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This leaves us with the limit as π₯ approaches zero of two sin of π₯ over two over π₯ is equal to one.
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And now we have a constant inside our limit, which is two.
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So we can factorize this constant out of our limit.
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And we simply divide both sides of the equation by two.
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The limit on the left of the equation here is looking very close to the limit weβre trying to evaluate.
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The only difference is that the fractions in the two limits are reciprocals of one another.
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In order to make these two limits identical, weβll be using the fact that the limit of a reciprocal is equal to the reciprocal of the limit.
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What this means is that the limit as π₯ approaches zero of π₯ over sin of π₯ over two is equal to one over the limit as π₯ approaches zero of sin of π₯ over two over π₯.
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We have just found this limit to be equal to one-half.
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So we can substitute this in here.
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And this gives us one over one-half, which is simply equal to two.
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And so here we have found the value of the limit which weβre trying to evaluate.
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And we can substitute this back into our equation.
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This tells us that the limit as π₯ approaches zero of sin π₯ over sin of π₯ over two is equal to one multiplied by two, giving us a solution of two.
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An alternative method to solve this question is to use a trigonometric identity.
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Weβll be using the fact that sin of two π is equal to two sin π cos π.
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If we let π equal π₯ over two, we obtain that the sin of π₯ is equal to two sin of π₯ over two timesed by cos of π₯ over two.
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And we can substitute this value of sin of π₯ into the numerator of our limit.
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In doing this, we obtain the limit as π₯ approaches zero of two sin of π₯ over two multiplied by cos of π₯ over two all over sin of π₯ over two.
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And so we can cancel the sin of π₯ over two in the top and bottom of the fraction.
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This leaves us with the limit as π₯ approaches zero of two cos of π₯ over two.
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And here we can simply apply direct substitution.
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And since cos of zero is equal to one, we obtain the same solution as earlier of two.
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In this last example, we saw how we can adapt the limit as π₯ approaches zero of sin of π₯ over π₯ is equal to one in order to show that the limit as π₯ approaches zero of π₯ over sin of π₯ over two is equal to two.
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Letβs consider the general case of the limit as π₯ approaches zero of sin of ππ₯ over π₯.
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Taking our original rule and substituting in ππ₯ for π₯, we obtain this.
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However, since π is a constant, if ππ₯ is approaching zero, then this must mean that π₯ is approaching zero.
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And so instead of writing ππ₯ is approaching zero, we can simply write π₯ is approaching zero since these two things are equivalent.
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Next, we can factorize out the π in the denominator of our fraction.
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Our fraction becomes one over π multiplied by sin of ππ₯ over π₯.
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Since one over π is just a constant, we can factorize it out of our limit.
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For our final step here, we multiply both sides by π.
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And this leaves us with a new rule.
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We have that the limit as π₯ approaches zero of sin of ππ₯ over π₯ is equal to π.
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We can further adapt this rule in order to find the limit as π₯ approaches zero of tan of ππ₯ over π₯.
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We start by writing tan of ππ₯ as sin of ππ₯ over cos of ππ₯.
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Next, weβll be using the fact that the limit of a product of functions is equal to the product of the limit of those functions.
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And we obtain this.
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And we can see that the limit on the left-hand side is equivalent to the rule we just derived.
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And so, therefore, itβs equal to π.
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And we can use direct substitution in order to find the limit on the right-hand side.
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Since cos of zero is just one, we obtain that this is equal to π.
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And so we found a new rule.
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Thatβs the limit as π₯ approaches zero of tan ππ₯ over π₯ is equal to π.
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Now weβre ready to move on to our next example.
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Find the limit as π₯ approaches zero of sin squared of seven π₯ plus three tan squared of three π₯ over eight π₯ squared.
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If we were to try direct substitution, we would obtain zero over zero, which is undefined.
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Letβs try to find this limit using these rules.
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Weβll also be using the fact that the limit of a sum of functions is equal to the sum of the limits of the functions.
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Hence, we can write our limit as the sum of these two limits.
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We notice that we can take a factor of one-eighth out of the first limit and a factor of three-eighths out of the second limit.
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Next, we notice that both numerators and both denominators are squares, enabling us to write our limits like this.
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Now we can use the fact that the limit of a square of a function is equal to the square of the limit of the function.
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In doing this, we are left with this.
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And we notice that our limits look very similar to the ones we wrote out at the start.
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Substituting π equals seven into the first limit rule, we see that our limit on the left must be equal to seven.
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And substituting π equals three into the second limit rule, we see that our limit on the right must be equal to three.
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We obtain one-eighth multiplied by seven squared plus three-eighths multiplied by three squared.
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Simplifying this down, we obtain a solution of 19 over two.
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Next, weβll be looking at a different rule which is useful for finding the limits of trigonometric functions of another form.
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The rule which weβll be using is the limit as π₯ approaches zero of one minus cos π₯ over π₯ is equal to zero.
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Now the proof of this rule is again beyond the scope of this video.
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However, we can think about this rule intuitively by considering the small angle approximation of cos.
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We have that, for small values of π₯, cos of π₯ is roughly equal to one minus π₯ squared over two.
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So as π₯ approaches zero, cos of π₯ will be approaching one minus π₯ squared over two.
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We obtain that the limit as π₯ approaches zero of one minus cos of π₯ over π₯ is roughly equal to the limit as π₯ approaches zero of one minus one minus π₯ squared over two all over π₯, which simplifies to the limit as π₯ approaches zero of π₯ over two.
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Using direct substitution, we see that this is equal to zero, which agrees with the rule which we stated at the beginning.
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Another way to see this intuitively is to consider the graph of one minus cos of π₯ over π₯.
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We can see from the graph that as the value of π₯ approaches zero, the line of the graph also approaches zero.
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We would see a similar result by using a table of values.
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Letβs now consider another example.
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Determine the limit as π₯ approaches zero of nine minus nine cos of seven π₯ over three π₯.
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First, we notice that we can cancel a factor of three from the top and bottom of this fraction, leaving us with this limit.
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Next, we notice that we can factor three out of the limit.
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If we attempt to do direct substitution at this point, we will see that we obtain zero over zero, which is undefined.
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Letβs instead use the fact that the limit as π₯ approaches zero of one minus cos of π₯ over π₯ is equal to zero.
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We can in fact adapt this formula in order to find that the limit as π₯ approaches zero of one minus cos of ππ₯ over π₯ is equal to zero, where π is just a constant.
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We can do this by substituting ππ₯ in for π₯ into the first rule.
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We obtain that the limit as ππ₯ goes to zero of one minus cos of ππ₯ over ππ₯ is equal to zero.
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Now since π is a constant and ππ₯ is going to zero, this means that π₯ is also going to zero.
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Instead of writing ππ₯ goes to zero, we can simply write π₯ goes to zero.
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Next, we notice we have a factor of π in the denominator of this fraction.
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And so we can factorize this π out of the limit.
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Next, we can multiply both sides of the equation by π.
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Since the right-hand side is zero, zero times π gives us zero.
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So the right-hand side remains as zero.
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And so now we have found that the limit as π₯ goes to zero of one minus cos of ππ₯ over π₯ is zero, which is what we were trying to show.
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By substituting π equals seven into this formula, we can see that our limit here is simply zero.
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And since three multiplied by zero is simply zero, we find that the solution here is simply zero.
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Next, letβs look at a slightly more tricky example.
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Find the limit as π₯ goes to π over two of two minus two sin of π₯ over four π₯ minus two π.
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First, we try solving it by direct substitution.
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However, weβll reach zero over zero, which is undefined.
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So we must find this limit by another means.
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Letβs start by cancelling a factor of two in both the numerator and denominator of the fraction.
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Now letβs consider some of the rules we know.
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When considering the first rule here, we notice that the value inside of the sine must be the same as the value in the denominator of the function.
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In our limit, we have sin of π₯ in the numerator.
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However, in the denominator, we have two π₯ minus π.
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And these two things are not equal.
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Therefore, we cannot use this first rule.
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In order to use the second rule, we require a cos π₯ in the numerator.
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However, in our limit, we currently have a sine.
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Here weβll be using an identity in order to change the sine to a cosine.
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We have that sin of π₯ is equal to cos of π₯ minus π by two.
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And we can substitute this into our limit.
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Factorizing the denominator of the fraction here, we can see that this is now of a very similar form to the rule which we know.
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At this stage, we need to perform a substitution.
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We will substitute in π’ is equal to π₯ minus π by two.
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However, we first need to consider what will happen to our limit.
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So thatβs π₯ approaching π by two.
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Well, weβll simply consider what happens to the value of π’ as π₯ approaches π by two.
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Our value of π’ will approach π by two minus π by two, which is simply zero.
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And so now weβre ready to substitute π’ is equal to π₯ minus π by two into our limit.
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We obtain the limit as π’ approaches zero of one minus cos of π’ over two π’.
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We can factor out the half.
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And now we notice that our limit is identical to our rule.
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And so, therefore, it must equal zero.
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And this gives us a solution of zero.
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In this last example, we saw how we have to be careful with trigonometric limits as sometimes it can be difficult to spot how to solve them.
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Itβs very important to keep the trigonometric identities in mind.
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Now letβs recap some of the key points of this video.
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Key Points.
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The limit as π₯ approaches zero of sin of π₯ over π₯ is equal to one.
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This gives us that the limit as π₯ approaches zero of sin of ππ₯ over π₯ is equal to π.
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And this then leads us on to the limit as π₯ approaches zero of tan of ππ₯ over π₯ is equal to π.
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We also have that the limit as π₯ approaches zero of one minus cos of π₯ over π₯ is equal to zero, which leads on to the limit as π₯ approaches zero of one minus cos of ππ₯ over π₯ is also equal to zero.
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And if we cannot solve a limit of a trigonometric function using direct substitution or one of the rules above, then we should try using some trigonometric identities, such as the ones shown here.