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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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Emperoraeneas
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  • Is there any specific reason for asking, or just curiosity? – egreg Apr 08 '19 at 21:36
  • Why do you want to do this? – GEdgar Apr 08 '19 at 21:36
  • 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 – Somos Apr 08 '19 at 21:44
  • 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 $ – Emperoraeneas Apr 08 '19 at 21:55
  • 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 – Emperoraeneas Apr 08 '19 at 21:57
  • If you assume those two things, then you can just break the limit up into a product of limits. After all, if $\lim_{x \to a} f(x)=L$ and $\lim_{x \to a}g(x)=M$, then $\lim_{x\to a}f(x)g(x)=LM$. – kccu Apr 08 '19 at 22:26
  • Can you please clarify? how do you break $\frac{ln(x+1)+1-e^x}{x^2}$ into a product – Emperoraeneas Apr 08 '19 at 22:36
  • You assumptions that $\lim_{x\to 0} \ln(x+1)/x = 1$ and $\lim_{x\to 0} (e^x-1)/x = 1$ are not sufficient to prove what y ou want to prove. You need to know more about $\ln$ and $\exp$. What algebraic manipulation is possible with only what you gave? – Somos Apr 09 '19 at 00:09

1 Answers1

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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 $L=1/2$.

Paramanand Singh
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