Video: Finding Limits Involving Trigonometric Functions

Find lim_(π‘₯ β†’ 0) ((tan 4π‘₯)/9π‘₯).

06:52

Video Transcript

Find the limit as π‘₯ approaches zero of the function tan of four π‘₯ divided by nine π‘₯.

The first thing we should try in any question asking us to find the limit like this is ask, can we use direct substitution? We can use direct substitution to find the limit of any function 𝑓 which is 𝑓 is a polynomial, trigonometric, exponential, or logarithmic function. If 𝑓 is the product or quotient of functions which we can evaluate by using direct substitution. Or if 𝑓 is the composition of two functions which we can evaluate by direct substitution.

We can attempt to justify using direct substitution on the function in our question in parts. We see that four π‘₯ is a polynomial, so this can be evaluated by using direct substitution. We then see that tan four π‘₯ is the composition of a trigonometric function and polynomial. And so, the numerator of the quotient in our question can be evaluated by using direct substitution. We also know that nine π‘₯ is a polynomial, and so the denominator of the quotient in our question can be evaluated by using direct substitution. Finally, since we have shown that the function in our question is the quotient of functions which we can evaluate by using direct substitution, we have justified our use of this tool.

So, we substitute the value of zero into our function to get that the limit, as π‘₯ approaches zero of tan four π‘₯ over nine π‘₯, is equal to tan of four multiplied by zero divided by nine multiplied by zero. We can simplify this to tan of zero over zero. We can evaluate tan of zero to just be equal to zero. So, this simplifies to zero divided by zero, which we know is an indeterminate form. Since direct substitution gave us an indeterminate form, we’re going to have to manipulate our expression of tan of four π‘₯ divided by nine π‘₯ into something we can work with. So, let’s clear some space to work with.

To solve this problem, we’re going to make use of the following well-known limit, the limit as πœƒ tends to zero of sin of πœƒ divided by πœƒ is equal to one. We can start by trying to rewrite our expression to be in terms of the sine function. We can do this by utilizing the tangent identity, which says that the tangent of πœƒ is equivalent to sin of πœƒ divided by cos of πœƒ. Since this identity is true for any value of πœƒ, we could replace πœƒ with four π‘₯, giving us that tan of four π‘₯ is equivalent to sin of four π‘₯ over cos of four π‘₯.

We can then use this to rewrite our limit as the limit as π‘₯ approaches zero of sin of four π‘₯ over cos π‘₯ all divided by nine π‘₯. We can simplify this further by using the fact that π‘Ž over 𝑏 all divided by 𝑐 is just equal to π‘Ž over 𝑏 multiplied by 𝑐. This gives us that the limit in our question is equal to the limit as π‘₯ approaches zero of sin of four π‘₯ divided by nine π‘₯ multiplied by the cos of four π‘₯. We can then rewrite our limit as the product of these two fractions.

Now, we also know that the limit of a product of two functions is equal to the product of their limits. We can use this to rewrite our limit as the limit as π‘₯ approaches zero of sin of four π‘₯ over nine π‘₯ multiplied by the limit as π‘₯ approaches zero of one over cos of four π‘₯. We know that one divided by the cos of four π‘₯ is the reciprocal of the composition of a trigonometric function and a polynomial. So, we can evaluate this limit by direct substitution.

This gives us one divided by the cos of four multiplied by zero, which we can evaluate to be one divided by one. Therefore, we can replace our limit with just one, giving us that the limit in our question is equal to the limit of sin four π‘₯ divided by nine π‘₯, as π‘₯ approaches zero. So, let’s clear some space so we can focus on this limit.

We can see that the limit we’re now trying to evaluate is very similar to the limit we have on the left, the limit as πœƒ approaches zero of sin of πœƒ over πœƒ is equal to one. The only differences are the coefficients and the fact that we have the variable πœƒ on the left and the variable π‘₯ on the right. Let’s replace our variable πœƒ with four π‘₯. This gives us that the limit as four π‘₯ approaches zero of sin of four π‘₯ over four π‘₯ is equal to one. We know that if four π‘₯ is tending to zero, then π‘₯ must also be tending to zero. So, this is the same as just saying π‘₯ tends to zero.

We can now see that the only difference between these two limits is the coefficient in the denominator. We want to change this coefficient of a nine into a four so that we can use the limit which we already know the value of. We can do this by noticing that nine π‘₯ is equal to nine over four multiplied by four π‘₯. This gives us the limit as π‘₯ tends to zero of sin of four π‘₯ over nine divided by four multiplied by four π‘₯. We can simplify this expression by multiplying both the numerator and the denominator by four over nine.

We get that the coefficients of nine over four and four over nine in the denominator will cancel. This gives us the limit as π‘₯ approaches zero of sin of four π‘₯ divided by four π‘₯ multiplied by four over nine. Now, we can use that the limit of a product is equal to the product of the limits to rewrite our limit as the limit as π‘₯ approaches zero of sin of four π‘₯ over four π‘₯ multiplied by the limit as π‘₯ approaches zero of four over nine.

We know that for any constant π‘˜, the limit as π‘₯ approaches π‘Ž of the constant π‘˜ is just equal to π‘˜ itself. So, we can evaluate the limit of the constant four over nine to just be equal to four over nine. And we have shown that the limit of sin of four π‘₯ over four π‘₯, as π‘₯ approaches zero, is equal to one. So, we can replace this limit with the value of one. This gives us that the limit of tan of four π‘₯ divided by nine π‘₯ as π‘₯ approaches zero is equal to four over nine.

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