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→∞ (tan⁻¹ x - π/2)/(1/x)

Accepted Answer

Welcome back, everyone. Apply Lapto's rules to find the limit of the following function as X approaches infinity F of X equals inverse of cotangent of X divided by 1 divided by X where given or answer choices A says -1, B1, C0, and D infinity. So let's go ahead and write down the limit. We are evaluating the limit as x approaches infinity. Of inverse of cotangent of X. And we are dividing that by the inverse of X, right? So, essentially what we want to do is begin with direct substitution. So we are going to take inverse of cotangent of infinity, and essentially when X approaches infinity, then the inverse of cotangent is going to approach 01 divided by X, well, when X approaches infinity, 1 divided by infinity is going to approach 0 as well. So we have an indeterminant form, 0 divided by 0, and this allows us to apply Laptal's rule, right? So we're going to take a limit as X approaches infinity, and according to Lapito's rule, we have to take the derivative of the numerator, inverse of cotangent of X. In the derivative of the denominator, so the derivative of a 1 divided by X, which is basically the derivative of X to the power of -1. Now let's recall that the derivative of the inverse of cotangent. is equal to -1 divided by 1 + X2, right? We have X, so there's no necessity to apply the chain rule. Now for the denominator, we are going to apply the power rule. So we bring down -1. This gives us the negative sign, and we decrease the original exponent by 1 unit. So the new exponent is -2. From here we're going to simplify the expression. Let's understand that negative X to the power of -2 is simply -1 divided by X2, right? So we can divide the two fractions. Which is going to give us X squared at the top. And 1 + x squared at the bottom, right? We're taking the first fraction and we're multiplying it by the reciprocal of the bottom fraction. So X2d goes on top, and we have canceled out the two negative signs. If we apply the right substitution once again, we're going to get infinity squared or basically infinity, divided by 1 plus infinity squared, which is infinity once again. And this is another form where we can apply Lapital's rule, right? So we have to apply it once again. We're going to differentiate x squared. In the numerator and we're going to differentiate. 1 + x 2 in the denominator, so it's the derivative of 1 plus the derivative of X2. And we know that the derivative of 1 is 0. Which leaves us with limit as X approaches infinity. Of the derivative of X2 divided by the derivative of X2, and we can already see that these can be canceled out. We can just show that each of those is 2 X and then cancel them out or just cancel them out immediately, right? So we're taking the limit of 1 which is independent of X. So we get limit as x approaches infinity of 1. And we get the same constant because it's independent of X. So the answer to this problem corresponds to the answer choice B, which is 1. Thank you for watching.

17–83. Limits Evaluate the following limits. Use l’Hôpital’s Rule when it is convenient and applicable.lim_x→∞ (tan⁻¹ x - π/2)/(1/x)

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

lim_x→∞ (tan⁻¹ x - π/2)/(1/x)