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# Prove that $\displaystyle \lim_{x \to 0^+} \ln x = -\infty$.

## Consider $\delta=e^{N} .0 < x < \delta \Rightarrow x < e^N \Longrightarrow \ln (x) < N .$ Therefore, by the definition of a limit, $\lim _{x \rightarrow 0^{+}} \ln (x)=-\infty$

Limits

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### Video Transcript

This is a problem. Number 43 of the Stuart Calculus ES edition section 2.4 prove that the limit has experts zero from the right of the function Ln of X is equal to negative infinity, and we're going to use a little bit of definition seven and definition for definition, for states for red hand limits and that there should exist adults a value such that between a X or between a and A plus steel toe. The function has a certain conditional, but in this case, this is an infinite limit. So from definition seven, what we have is that we want to choose a value, and that is positive that if this is true, then this function is always going to be less than M. So this is a bit of a combination between definitions four and seven. So we're gonna do is we're going to take first. We're going to understand what this is in our case, A zero since we're approaching zero from the right. Therefore, we're in this area between X zero and delta the function it is olympics less than in. And if we wanted to choose an appropriate ex, we take maybe the solve for X. We'd take the exponential to both sides. And by comparing these two cases here, we see that an appropriate choice of Delta in order to be consistent with this definition of a limit is e to the power capital in so the states that for any end, we are able to find a delta that is positive and is consistent with these conditions, Therefore, proving that this limit is equal to negative infinity.

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