# Calculate this limit without L'Hospital's rule or series expansion [duplicate]

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 $$0$$:

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

## marked as duplicate by Somos, Shailesh, José Carlos Santos, lab bhattacharjee limits StackExchange.ready(function() { if (StackExchange.options.isMobile) return; $('.dupe-hammer-message-hover:not(.hover-bound)').each(function() { var$hover = $(this).addClass('hover-bound'),$msg = $hover.siblings('.dupe-hammer-message');$hover.hover( function() { $hover.showInfoMessage('', { messageElement:$msg.clone().show(), transient: false, position: { my: 'bottom left', at: 'top center', offsetTop: -7 }, dismissable: false, relativeToBody: true }); }, function() { StackExchange.helpers.removeMessages(); } ); }); }); Apr 11 at 16:36

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• Is there any specific reason for asking, or just curiosity? – egreg Apr 8 at 21:36
• Why do you want to do this? – GEdgar Apr 8 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 8 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 8 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 8 at 21:57

## 1 Answer

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 $$L=1/2$$.

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