So, I have to solve the limit problem using a derivative.
The problem is as follows: $$\lim_{x\to0}\frac{(1+x)^{1/x}-e}{x}$$
I really don't know what to do. Can someone help?
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5I think you want L'Hopital's rule. – bob.sacamento Apr 29 '19 at 19:26
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1really ? what about starting by writing $a^b=\exp(b\ln(a))$. if you do nothing, sure, it won't solve by itslef. Then use expansion of $\ln$ around $1$. – zwim Apr 29 '19 at 19:27
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Calculate an asymptotic expansion of $(1+x)^{1/x}$ at order $1$. – Bernard Apr 29 '19 at 19:31
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Hint :
Observe that :
$$\lim_{x \to 0} (1+x)^{1/x} = \lim_{x\to 0} \left(1 + \frac{1}{\frac{1}{x}} \right)^{\frac{1}{x}} = e$$
Thus, the numerator of your fraction inside the limit tends to zero, while the same holds for the denominator.
Use L'Hopital's rule now and work on the calculation to derive the result.
Note : In case the entry line seemed weird to you, recall the sequential definition of $e$, as :
$$\lim_{n \to \infty} \left( 1 + \frac{1}{n}\right)^{n} = e$$
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1I think the sequential definition of $e$ is $e:=\lim\limits_{n \to \infty} \left(1+\frac{1}{n}\right)^{n}$. – Botond Apr 29 '19 at 19:45
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