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Evaluate each limit (if it exists). Use $L$ Hospital's rule (if appropriate).$$\lim _{x \rightarrow \infty} \frac{1+e^{2 x}}{2+\ln x}$$

$\infty$

Calculus 1 / AB

Chapter 27

Differentiation of Transcendental Functions

Section 7

L'Hospital's Rule

Derivatives

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So we're looking at the limit as X goes to infinity of one plus E. To the two X over two plus natural log effects. And uh if you look at the graphs of those two functions, like shifting up doesn't really matter. Um And the expo doesn't really matter. So that's the numerator, it's definitely going I think it's definite that it's going to infinity if you look at the graph. Uh and then same thing with two plus natural log. So it just shifts the graph up. But the denominator is still going to infinity as well. So we have this indeterminant form of infinity over infinity to do low petals rule. Um Yeah. So how do we do love details rules? So he leaves the limit the same and we take the derivative of the top. So the heat of the two X. Because the derivative of one is zero. And then don't forget the chain rule. You have to take the derivative of two X. Which is to over the derivative of natural log of X would be one over X. Um So as we're looking at this problem, the numerator is still going to infinity. Going infinitely up. But the graph of one over X may not do this in a different color as we move to the right, we're getting closer and closer to zero. So this is no longer indeterminate form. Um But as far as the uh well maybe I shouldn't think of it this way. Maybe what I should be thinking about is instead of dividing by one over X, we can multiply by the reciprocal. So that function would be the same thing as two X. E. To the two X. Yeah, this is a better way of thinking about it because now we're taking infinity and multiplying by another infinity. So we are going infinitely up. I'm just not a fan of saying it equals infinity because that's not a number and limits need to equal a number. You might have a teacher that's okay with that as your answer, but I would rather see does not exist Danny, but this is correct as well.

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