In Exercises 1–6, use l’Hôpital’s Rule to evaluate the limit. ...

Question

In Exercises 1–6, use l’Hôpital’s Rule to evaluate the limit. Then evaluate the limit using a method studied in Chapter 2.

3. lim (x → ∞) (5x² - 3x) / (7x² + 1)

Accepted Answer

Welcome back everyone. Use Slappi's rule to evaluate the limit. Limit as X approaches negative infinity of 5 X minus 6 X 2 divided by 3 X2 minus 2. Let's begin rewriting the limit. Limit as X approaches negative infinity. 5 X minus 6 X 2 divided by 3 X 2 minus 2. Remember that the first step before we apply Lapto's rule is to check if we have the correct and determinant form. So, if we substitute infinity for X, we're going to get negative infinity. -6 multiplied by negative infinity squad is minus infinity. Squared, which is basically minus infinity. And then we are dividing by. We multiplied by negative infinity squared, that's positive infinity minus 2, well, it basically leaves us with infinity in the denominator. So now notice that we're getting a form. Negative infinity minus infinity basically gives us negative infinity divided by positive infinity, right? We can basically rewrite it. That's negative infinity divided by infinity, and this is one of the indeterminate forms that we can use to apply Labital's rule. So, what we're going to do from here is Basically differentiate the numerator and the denominator, right? This is what Lapito's rule says. Let's begin by finding the derivative of 5 X minus 6 X2, and then we will find the derivative of 3 X2 minus 2. So, for the numerator, we can simply apply the power rule to get. 5 minus 12 X, and in the denominator, we're going to get 6 X. Minus 0, right? The derivative of a constant is 0. According to the rule, our next step is. To again, substitute X approaches infinity, we're going to get 5 minus 12 multiplied by negative infinity. So that's basically 5+ infinity, which is positive infinity. We can just write infinity in the numerator, and for the denominator, we're going to get 6 multiplied by negative infinity, which is negative infinity. So we still have the same in the terminal form, right? Meaning we're going to apply lapto's rule once again, let's find each derivative. For the numerator, the derivative of 5 is 0. The derivative of X is 1, so we basically get -12. And we're dividing by the derivative of the denominator. The derivative of X is 1, so we basically get 6. Now, notice that we're taking the limit of a constant, right? Limit as X approaches negative infinity of -2. The limit of a constant is equal to the constant itself, meaning the correct answer is -2. Thank you for watching.

In Exercises 1–6, use l’Hôpital’s Rule to evaluate the limit. Then evaluate the limit using a method studied in Chapter 2.3. lim (x → ∞) (5x² - 3x) / (7x² + 1)

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In Exercises 1–6, use l’Hôpital’s Rule to evaluate the limit. Then evaluate the limit using a method studied in Chapter 2.

3. lim (x → ∞) (5x² - 3x) / (7x² + 1)