WEBVTT
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Find the limit as π₯ approaches zero of six π₯ cotan squared of four π₯ all divided by the cosec of eight π₯.
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The question is asking us to find the limit of a combination of trigonometric functions.
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And we see if we were to try to use direct substitution, since our limit is as π₯ is approaching zero, we have that the cotan of four times zero is undefined.
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And since this is undefined, we canβt use direct substitution in this case.
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So, weβre going to need to manipulate this expression and so we can use direct substitution or where we can use one of our rules for trigonometric limits.
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Weβre going to start by rewriting this expression entirely in terms of the sine and cosine function.
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We recall taking the cotan of π is equivalent to one divided by the tan of π, which is the same as saying the cos of π divided by the sin of π.
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However, in our expression, weβre taking the cotan of four π₯, so weβll replace π with four π₯.
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This gives us the cotan of four π₯ is equivalent to the cos of four π₯ divided by the sin of four π₯.
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Next, we want to rewrite the cosec of eight π₯ in terms of the sine and cosine function.
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We can do this by recalling the cosec of an angle π is equivalent to one divided by the sin of π.
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And as we did before, since we want the cosec of eight π₯, weβll replace π with eight π₯.
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This gives us the cosec of eight π₯ is equivalent to one divided by the sin of eight π₯.
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Replacing the cotan of four π₯ with the cos of four π₯ divided by the sin of four π₯ gives us a new numerator in our limit of six π₯ multiplied by the cos of four π₯ over the sin of four π₯ squared.
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Then, replacing the cosec of eight π₯ with one divided by the sin of eight π₯ gives us a new denominator in our limit of one divided by the sin of eight π₯.
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Now, instead of dividing by the fraction one divided by the sin of eight π₯, we can multiply it by the reciprocal.
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So, multiplying by the reciprocal of one divided by the sin of eight π₯ and expanding our square over our parentheses gives us that our limit is equal to the limit as π₯ approaches zero of six π₯ times cos squared four π₯ times sin of eight π₯ all divided by the sin squared of four π₯.
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However, we still canβt use direct substitution on this limit.
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Since our limit is as π₯ is approaching zero, weβll get a denominator of the sin squared of four times zero, which is zero.
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And our numerator will have the sin of eight times zero, which is zero, giving us an indeterminate form.
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So, we still need to perform more manipulations to evaluate this limit.
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Weβll start by recalling the double angle formula for sine, which tells us the sin of two π is equivalent to two sin π cos π.
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We want to apply this to the sin of eight π₯ in our numerator.
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So, we will end up with the sin of four π₯ in our numerator, which we can cancel with one of the factors of sin of four π₯ in the denominator.
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So, by setting π equal to four π₯, we have the sin of eight π₯ is equivalent to two sin of four π₯ cos of four π₯.
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So, by substituting sin of eight π₯ is equal to two sin four π₯ cos of four π₯ into our limit, we get a new expression for our limit.
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And we can cancel the shared factor of sin of four π₯ in our numerator with one of the factors in our denominator.
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Simplifying this gives us that our limit is equal to the limit as π₯ approaches zero of 12π₯ cos cubed of four π₯ all divided by the sin of four π₯.
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However, we still canβt perform direct substitution on this limit.
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Since our limit is as π₯ is approaching zero in our numerator, our factor of π₯ will approach zero.
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And in our denominator, the sin of four π₯ will approach the sin of four times zero, which is zero, giving us an indeterminate form.
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So, we still need to perform more manipulation.
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Weβll do this by using one of our standard limit results for trigonometric functions.
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Weβll use the fact that the limit as π₯ approaches zero of the sin of π₯ divided by π₯ is equal to one.
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This is a standard result which we should commit to memory.
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However, in our limit we have π₯ in the numerator and our sine function in the denominator.
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So, weβll need to take the reciprocal of this limit.
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We recall the reciprocal of a limit is equal to the limit of the reciprocal.
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So, the limit as π₯ approaches zero of π₯ divided by the sin of π₯ is equal to the reciprocal of one, which is just equal to one.
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Now, we see that weβre taking the sin of four π₯ in our limit.
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So, weβll need to replace π₯ with four π₯.
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This gives us the limit as four π₯ approaches zero of four π₯ divided by the sin of four π₯ is equal to one.
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And if four π₯ is approaching zero, then π₯ is approaching zero.
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So, we have the limit as π₯ approaches zero of four π₯ divided by the sin of four π₯ is equal to one.
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So, we just need a four π₯ in our numerator.
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And we can notice that 12π₯ is equal to three multiplied by four π₯.
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This gives us our limit is equal to the limit as π₯ approaches zero of three times four π₯ over sin of four π₯ all multiplied by the cos cubed of four π₯.
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Since three is a constant, we can just take it outside of our limit.
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Next, weβll use the limit of a product is equal to the product of a limit to split our limit into the product of two limits.
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This gives us three multiplied by the limit as π₯ approaches zero of four π₯ over sin π₯ times the limit as π₯ approaches zero of the cos cubed of four π₯.
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Weβre now almost ready to evaluate this limit.
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Weβll just change the limit of the cos cubed of four π₯ using the power rule for limits to be the limit of the cos of four π₯ all cubed.
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We have the limit as π₯ approaches zero of four π₯ divided by sin π₯ is equal to one.
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And we have the limit as π₯ approaches zero of the cos of π₯ is also equal to one.
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So, our limit is equal to three multiplied by one multiplied by one cubed, which is equal to three.
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Therefore, weβve shown the limit as π₯ approaches zero of six π₯ multiplied by the cotan squared of four π₯ all divided by the cosec eight π₯ is equal to three.