17–83. Limits Evaluate the following limits. Use l’Hôpital’s R...

Question

17–83. Limits Evaluate the following limits. Use l’Hôpital’s Rule when it is convenient and applicable.

lim_x→ π/2⁻ (tanx ) / (3 / (2x - π))

Accepted Answer

Hi everyone. Let's take a look at this practice problem. This problem says to apply Lopital's rule to find the limit of the following function as x approaches 3 pi divided by 2 from the left. When we go to the function F of X is equal to the tangent of X divided by the quantity of 5 divided by the quantity of 2 divided by 3 multiplied by X minus pi in quantity. We have 4 possible choices as our answers. For choice A, we have minus 2 divided by 15. For choice B, we have 2 divided by 15. For choice C, we have minus 2 divided by 5, and for choice D, we have 2 divided by 5. Now, we're asked to apply Lopital's rule to find our limit here. And the first step that we actually want to do is to rewrite our function F of X. So we'll have F of X is equal to, and we're gonna rewrite tangent of X as the sign of X. Divided by cosine of X. And then we're going to multiply this by the reciprocal of the fraction in the denominator of our original function. So this is going to be multiplied by the quantity of 2 divided by 3, multiplied by X minus pi. In quantity divided by 5. And so simplified in this expression, this is going to be equal to. Sign of X Multiplied by the quantity of 2 divided by 3, multiplied by X minus pi in quantity, divided by the quantity of 5 multiplied by cosine of X. So now, if we take the limit, Of as X approaches, 3 pi divided by 2 from the left of F of X. This is going to be equal to the limit. As x approaches, 3 pi divided by 2 from the left. Of the quantity of sine of x. Multiplied by The quantity of 2 divided by 3, multiplied by X minus pi in quantity, divided by. 5 multiplied by cosine of X. So if we apply direct substitution. In order to evaluate this limit, we are actually going to get this limit to be equal to 0 divided by 0. Since in the numerator, we will have 3 divided by 2 divided by 3, multiplied by 3 pi divided by 2 minus pi, which gives me 0, as well as having cosine of 3 pi divided by 2, which is equal to 0 in the denominator. So we'll get 0 divided by 0, which is the indeterminate form. So we're going to apply Lobital's rule. So that means that we're going to have the limit. As X approaches. 3 pi divided by 2 from the left of F of X. It's going to be equal to the limit. As x approaches 3 pi divided by 2 from the left of the quantity, and here we'll take a derivative with respect to X of both our numerator and denominator. So that's what we do when we apply Lopital's rules. So we'll have the derivative with respect to X. Of the quantity of sin of x. Multiplied by the quantity of 2 divided by 3, multiplied by X minus pi in quantity, and that is divided by the derivative with respect to X. Of the quantity of 5 multiplied by cosine of X in quantity. So, taking the derivative, this is going to be equal to the limit. As X approaches 3 pi divided by 2 from the left of the quantity, and in the numerator we'll have to apply our product rules, so we'll take the derivative of the first term, which is the derivative of sine of X, which is cosine of X. That gets multiplied by the second term, which is the quantity of 2 divided by 3, multiplied by X minus pi in quantity. Then we'll have plus the first term, which is sin of X. Multiplied by the derivative of the second term, so that's gonna be the derivative with respect to x of the quantity of 2 divided by 3, multiplied by X minus pi. When I take that derivative, I get 2 divided by 3. And this whole quantity is going to be divided by The derivative with respect to X of 5 multiplied by cosine of X, which gives me minus 5 multiplied by sin of x. So, at this point, we can apply direct substitution to evaluate this limit. So this is going to be equal to. Cosine of 3 pi divided by 2. Multiplied by the quantity of 2 divided by 3, multiplied by the quantity of 3 pi divided by 2 in quantity minus pi in quantity, plus 2 divided by 3 multiplied by the sign of the quantity of 3 pi divided by 2 in quantity. That whole quantity is divided by. The quantity of minus 5 multiplied by the sign of 3 pi divided by 2. So evaluate in this expression, this is going to be equal to. For cosine of 3 pi divided by 2, that is 0. That gets multiplied by 0 again, since 2 divided by 3, multiplied by 3 pi divided by 2 minus pi is also 0. Then we will have plus. 2 divided by 3, multiplied by minus 1, since the sign of 3 pi divided by 2 is -1, and this whole quantity is divided by. -5, multiplied by -1. So simplify this expression, this is going to be equal to. Minus 2 divided by 15. So this is the answer that we're looking for for this problem. And when we compare our answer to our answer choices, we see that the correct answer choice corresponds to answer choice A. So I hope that this explanation has been useful, and I'll see you in the next video.

17–83. Limits Evaluate the following limits. Use l’Hôpital’s Rule when it is convenient and applicable.lim_x→ π/2⁻ (tanx ) / (3 / (2x - π))

Watch next

17–83. Limits Evaluate the following limits. Use l’Hôpital’s Rule when it is convenient and applicable.

lim_x→ π/2⁻ (tanx ) / (3 / (2x - π))