Problem

The accompanying schematic diagram represents an …

02:22

Problem 62 Hard Difficulty

Verify that L'Hópital's rule is of no help in finding the limit; then find the limit, if it exists, by some other method.
$$\lim _{x \rightarrow+\infty} \frac{x(2+\sin x)}{x^{2}+1}$$

Answer

0

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Calculus 1 / AB

Calculus Early Transcendentals

Chapter 3

TOPICS IN DIFFERENTIATION

Section 6

L Hopital's Rule; Indeterminate Forms

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Video Transcript

we're starting with limit as X goes to infinity of X times two plus sine X, divided by X squared plus one. Now notice that the numerator here is tending to positive infinity. Okay, it's it's tending to positive infinity because we know that the thing in the parentheses here is always is always positive. So the numerator is tending to positive infinity, the denominators tending the positive infinity. That's an indeterminant form of infinity over infinity. So we would expect the blue petals rule would apply. But remember that Lou Petals rule only applies if, after we apply it, we get a limit that exists. Okay, so we apply it and get this, then the other half of the product room by two X. Okay. And now notice No. Now notice that we could split this up into one over x sign of X over X plus co sign of X divided by two. And that gives us zero plus one plus this limit as X goes to infinity of co sign X divided by two. But this thing does not exist because it oscillates Okay, So because this limit does not exist, lo petals rule does not tell us anything about this limit. Let's go ahead and see if we we can't calculate its value. We know that this is actually going to rebounded above by two plus one Here, divide by X squared plus one and we're going to bound below by x times two minus one, divided by X squared, plus one. Okay. And now, um, it's a quick check that the limit on the right is equal to zero. We could, for example, just eyeball it and see that the that the polynomial the numerator has degree less than the degree and the denominator. Or we could apply loop atolls rule. Similarly, choose either arguments. We get that the lift and heights left hand limit. Excuse me? The limit on the left hand side is equal to zero, which tells us by the squeeze serum that this limit that we're looking for is equal to zero, and we're done

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