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Find the limits.$$\lim _{x \rightarrow+\infty}\left[x-\ln \left(x^{2}+1\right)\right]$$
$$\lim _{x \rightarrow+\infty}\left[x-\ln \left(x^{2}+2\right)\right]=+\infty$$
Calculus 1 / AB
Chapter 3
TOPICS IN DIFFERENTIATION
Section 6
L Hopital's Rule; Indeterminate Forms
Functions
Limits
Derivatives
Differentiation
Continuous Functions
Applications of the Derivative
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So we're starting with a limit as X goes to infinity of X minus natural log X squared plus one. And if we just try to evaluate this limit, we noticed that we get infinity minus infinity. Okay. In particular, this natural log X squared plus one is going to infinity because the X squared plus ones going to infinity. And so the natural log of that is going to So we haven't indeterminant for infinity minus infinity. We'll need to somehow rewrite this as a fraction in order to get anywhere. So let's go ahead and rewrite. The X is a fraction or excuse me, rewrite X using natural log so we can rewrite it as natural log of eat the X And now we can go ahead and use our log properties, and here we can go ahead. And because natural log is a continuous function, we can interchange the natural log in the limit sign. So we've natural log with limit of X to infinity e to the X, divided by X squared plus one. And now here, if we try to take this limit on the on the inside of the parentheses, um, we end up getting infinity over infinity. But it's since it's a fraction we can go ahead and use low petals rule. Okay, we're in the same sort of situation. Infinity over infinity. Let's go ahead and use low petals rule. And here we see that the limit on the inside is now going to infinity. So our answer here as infinity, that's it.
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