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

Find lim_(π‘₯ β†’ πœ‹/6) (9(πœ‹ βˆ’ 6π‘₯))/(tan 6π‘₯).

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Video Transcript

Find the limit as π‘₯ approaches πœ‹ by six of nine times πœ‹ minus six π‘₯ all divided by the tan of six π‘₯.

In this question, we’re asked to evaluate the limit of the quotient of a linear function and a trigonometric function. And since we can evaluate both of these limits by using direct substitution, we can attempt to evaluate this limit by using direct substitution. Substituting π‘₯ is equal to πœ‹ by six into our function, we get nine multiplied by πœ‹ minus six times πœ‹ by six all divided by the tan of six times πœ‹ by six. And six times πœ‹ by six is just equal to πœ‹, so our numerator simplifies to give us nine times πœ‹ minus πœ‹, which is zero. And our denominator simplifies to give us the tan of πœ‹, which is also zero. Therefore, trying to evaluate this limit by direct substitution gives us the indeterminate form zero divided by zero.

So we’re going to need to evaluate this limit using a different method. And since this is the limit of the quotient of a linear function and a trigonometric function, one way of doing this is to recall one of our trigonometric limit results. We recall for any real constant π‘Ž, the limit as π‘₯ approaches zero of the tan of π‘Ž times π‘₯ all divided by π‘₯ is equal to π‘Ž. We can’t just directly apply this limit result for a few reasons. First, we’re not taking the limit as π‘₯ approaches zero; we’re taking the limit as π‘₯ approaches πœ‹ by six. Second, we have tan of six π‘₯ in our denominator. However, in our limit result, the tan of π‘Žπ‘₯ is in the numerator.

However, we can fix this in a few steps. First, to get tan of six π‘₯ in the denominator of our result, we need to take the reciprocal of both sides of the equation. Taking the reciprocal of both sides of our limit result and applying the power rule for limits, we get the limit as π‘₯ approaches zero of π‘₯ divided by the tan of π‘Žπ‘₯ is equal to one divided by π‘Ž for any real constant π‘Ž not equal to zero. However, we’re still taking the limit as π‘₯ approaches zero. And in the limit we’re asked to evaluate, we’re taking the limit as π‘₯ approaches πœ‹ by six. And we can fix this by using a substitution. We’ll set 𝑦 equal to π‘₯ minus πœ‹ by six.

Now, as our values of π‘₯ get closer and closer to πœ‹ by six, π‘₯ minus πœ‹ by six will get closer and closer to zero. So our values of 𝑦 will approach zero. We now want to substitute this into our limit result. To do this, we’re going to need to find an expression for π‘₯ in terms of 𝑦. And we can do this by adding πœ‹ by six to both sides of our substitution. We get that π‘₯ is equal to 𝑦 plus πœ‹ by six. Substituting this into our limit result, we get the limit as 𝑦 approaches zero of nine multiplied by πœ‹ minus six times 𝑦 plus πœ‹ by six all divided by the tan of six times 𝑦 plus πœ‹ by six.

And we can simplify this. First, we’ll start by distributing negative six over the parentheses in our numerator and six over the parentheses in our denominator. This gives us the limit as 𝑦 approaches zero of nine times πœ‹ minus six 𝑦 minus πœ‹ divided by the tan of six 𝑦 plus πœ‹. And we can simplify this further. In our numerator, we have πœ‹ minus πœ‹, which is equal to zero. And we can also simplify the denominator by recalling the tangent function has a period of πœ‹. So the tan of six 𝑦 plus πœ‹ will be equal to the tan of six 𝑦.

Finally, to match our limit result, we want a constant factor of one in the numerator, so we need to take the factor of nine outside of our limit and the factor of negative six. This then leaves us with negative 54 multiplied by the limit as 𝑦 approaches zero of 𝑦 divided by the tan of six 𝑦. And we can directly evaluate this limit by using our limit result. The value of π‘Ž is six, so it evaluates to give us one over six. And so replacing this with one over six, we get negative 54 over six, which we can evaluate is negative nine, which is our final answer. Therefore, we were able to show the limit as π‘₯ approaches πœ‹ by six of nine times πœ‹ minus six π‘₯ divided by the tan of six π‘₯ is negative nine.

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