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This question already has an answer here:

Calculating the limit as $x$ approaches $0$ using L'Hospital's rule or series expansion is straightforward, but how to evaluate the limit without either of those techniques.

How to calculate the following limit as $x$ approaches $0$:

$\dfrac{\ln(x+1)+1-e^x}{x^2}$

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marked as duplicate by Somos, Shailesh, José Carlos Santos, lab bhattacharjee limits Apr 11 at 16:36

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

  • $\begingroup$ Is there any specific reason for asking, or just curiosity? $\endgroup$ – egreg Apr 8 at 21:36
  • $\begingroup$ Why do you want to do this? $\endgroup$ – GEdgar Apr 8 at 21:36
  • $\begingroup$ Okay, so exactly what are we allowed to know about the properties of $\ln(x)$ and $\exp(x)$? You can't expect an answer until we know the answer to this question $\endgroup$ – Somos Apr 8 at 21:44
  • $\begingroup$ I suppose we can assume $\lim_{x \to 0} \frac{ln(x+1)}{x} = 1 $ and $\lim_{x \to 0} \frac{e^x-1}{x} = 1 $ $\endgroup$ – Emperoraeneas Apr 8 at 21:55
  • $\begingroup$ I am asking this question to see whether it is possible and what insights can be used to evaluate such a limit using algebraic manipulation $\endgroup$ – Emperoraeneas Apr 8 at 21:57
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This question boils down to showing that the limit $$L=\lim _{x\to 0}\frac{e^x-1-x}{x^2}$$ exists. Using $e^x-1=t$ we can see that the above implies that $$L=\lim_{x\to 0}\frac{x-\log(1+x)}{x^2}$$ and adding this to the first limit we get the desired limit in question as $-2L$. You can use binomial theorem and the definition $$e^x=\lim_{n\to\infty} \left(1+\frac{x}{n}\right)^n$$ to get $L=1/2$.

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  • $\begingroup$ Unless I'm missing something, using $e^t-1=t$ only implies $L = \lim_{t \to 0} \frac{t-ln(t+1)}{(ln(t+1))^2}$ $\endgroup$ – Emperoraeneas Apr 9 at 18:04
  • $\begingroup$ @Emperoraeneas: but you can replace denominator by $t^2$ by multiplying the above limit with $\lim_{t\to 0}\left(\dfrac{\ln(1+t)}{t}\right)^2=1$. $\endgroup$ – Paramanand Singh Apr 10 at 1:44
  • $\begingroup$ Makes sense, thanks! $\endgroup$ – Emperoraeneas Apr 10 at 2:58

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