Evaluate the following limits in two different ways: with and ...

Question

Evaluate the following limits in two different ways: with and without l’Hôpital’s Rule.

lim_x→∞ (2x⁵ - x + 1) / (5x⁶ + x)

Accepted Answer

Evaluate the limit using Hobbit's rule, where we have the limit as x approaches infinity of 4x4 minus 5 x + 7 divided by 3 x 5th plus X. We have 4 possible answers being negative 4/3, 0, 3/4, or 4/3. Now, to solve this, let's first check if this meets the criteria for the Ho's rule. We have the limit, as X approaches infinity. Now, if we were to plug in Infinity, we end up getting the form. Of infinity divided by infinity. Abital's rule works. If you have the form infinity divided by infinity or 0 divided by 0. Because we have this form, you can take the rule. This rule states that If we have two functions, F divided by G. We can find the limit By taking the limit As x approaches whatever value. Of their derivatives F X divided by G of X. So now, let's take our derivatives. Our first derivative. We have 4x to the 4th, that goes to 16 X to the 3rd. Derivative of 5 X goes to 5. Leaving us 16 X 3 minus 5. This is divided by 15, X to the 4th, plus 1. Now, if we were to check our limit here about plugging in infinity, we noticed that this form is still infinity divided by infinity. Because that's the case, we can take Lapi's rule again. We have 48 to the second. Divided by 60. X to the 3rd. Now we can simplify this. To get 48 divided by 60 X. Or or divided by 5 X. Now, if we take the limit, as X approaches infinity, of 4 divided by 5 X. We end up getting 4 divided by infinity. Which approaches 0. This means the limit to our problem is 0, making this answer B. OK, I hope to help you solve the problem. Thank you for watching. Goodbye.

Evaluate the following limits in two different ways: with and without l’Hôpital’s Rule.lim_x→∞ (2x⁵ - x + 1) / (5x⁶ + x)

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Evaluate the following limits in two different ways: with and without l’Hôpital’s Rule.
lim_x→∞ (2x⁵ - x + 1) / (5x⁶ + x)