Question Video: Finding Limits Involving Trigonometric Functions Mathematics • Higher Education

Find lim_(π‘₯ β†’ 0) ((6π‘₯ cotΒ² 4π‘₯ ) / csc 8π‘₯).

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Video Transcript

Find the limit as π‘₯ approaches zero of six π‘₯ cotan squared of four π‘₯ all divided by the cosec of eight π‘₯.

The question is asking us to find the limit of a combination of trigonometric functions. 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. And since this is undefined, we can’t use direct substitution in this case. So, we’re 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.

We’re going to start by rewriting this expression entirely in terms of the sine and cosine function. We recall taking the cotan of πœƒ is equivalent to one divided by the tan of πœƒ, which is the same as saying the cos of πœƒ divided by the sin of πœƒ. However, in our expression, we’re taking the cotan of four π‘₯, so we’ll replace πœƒ with four π‘₯. This gives us the cotan of four π‘₯ is equivalent to the cos of four π‘₯ divided by the sin of four π‘₯.

Next, we want to rewrite the cosec of eight π‘₯ in terms of the sine and cosine function. We can do this by recalling the cosec of an angle πœƒ is equivalent to one divided by the sin of πœƒ. And as we did before, since we want the cosec of eight π‘₯, we’ll replace πœƒ with eight π‘₯. This gives us the cosec of eight π‘₯ is equivalent to one divided by the sin of eight π‘₯.

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. 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‘₯.

Now, instead of dividing by the fraction one divided by the sin of eight π‘₯, we can multiply it by the reciprocal. 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‘₯.

However, we still can’t use direct substitution on this limit. Since our limit is as π‘₯ is approaching zero, we’ll get a denominator of the sin squared of four times zero, which is zero. And our numerator will have the sin of eight times zero, which is zero, giving us an indeterminate form. So, we still need to perform more manipulations to evaluate this limit.

We’ll start by recalling the double angle formula for sine, which tells us the sin of two πœƒ is equivalent to two sin πœƒ cos πœƒ. We want to apply this to the sin of eight π‘₯ in our numerator. 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. So, by setting πœƒ equal to four π‘₯, we have the sin of eight π‘₯ is equivalent to two sin of four π‘₯ cos of four π‘₯.

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. And we can cancel the shared factor of sin of four π‘₯ in our numerator with one of the factors in our denominator. 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‘₯.

However, we still can’t perform direct substitution on this limit. Since our limit is as π‘₯ is approaching zero in our numerator, our factor of π‘₯ will approach zero. 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. So, we still need to perform more manipulation. We’ll do this by using one of our standard limit results for trigonometric functions.

We’ll use the fact that the limit as π‘₯ approaches zero of the sin of π‘₯ divided by π‘₯ is equal to one. This is a standard result which we should commit to memory. However, in our limit we have π‘₯ in the numerator and our sine function in the denominator. So, we’ll need to take the reciprocal of this limit. We recall the reciprocal of a limit is equal to the limit of the reciprocal. 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.

Now, we see that we’re taking the sin of four π‘₯ in our limit. So, we’ll need to replace π‘₯ with four π‘₯. This gives us the limit as four π‘₯ approaches zero of four π‘₯ divided by the sin of four π‘₯ is equal to one. And if four π‘₯ is approaching zero, then π‘₯ is approaching zero. So, we have the limit as π‘₯ approaches zero of four π‘₯ divided by the sin of four π‘₯ is equal to one.

So, we just need a four π‘₯ in our numerator. And we can notice that 12π‘₯ is equal to three multiplied by four π‘₯. 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‘₯. Since three is a constant, we can just take it outside of our limit.

Next, we’ll use the limit of a product is equal to the product of a limit to split our limit into the product of two limits. 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‘₯. We’re now almost ready to evaluate this limit. We’ll 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.

We have the limit as π‘₯ approaches zero of four π‘₯ divided by sin π‘₯ is equal to one. And we have the limit as π‘₯ approaches zero of the cos of π‘₯ is also equal to one. So, our limit is equal to three multiplied by one multiplied by one cubed, which is equal to three. Therefore, we’ve 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.

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