Evaluate $$\lim_{x\to0} \frac{(1+x)^{1/x}-e}{x}$$
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4What are your thoughts on the problem? – John Jan 15 '15 at 18:54
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1i cannot understand how to solve – rickriordan Jan 15 '15 at 18:58
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2use the rules of L'Hospital – Dr. Sonnhard Graubner Jan 15 '15 at 18:59
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1i am using them but no answer is coming – rickriordan Jan 15 '15 at 19:01
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This was already asked and answered just couple of days ago. Some university's Mathematics Dept. must have asked this question in some weekly exercise sheet, test or something. – Timbuc Jan 15 '15 at 19:16
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See also: How to solve this limit: $\lim\limits_{x\to0}\frac{(1+x)^{1/x}-e}x$? – Martin Sleziak Dec 06 '19 at 06:59
1 Answers
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We know that $\lim_{x\to0}\color{red}{\frac{e^x-1}{x}}=1$, or we prove it using L'Hospital.
We know that $\lim_{x\to0}\frac{\ln(1+x)}{x}=1$, or we prove it using L'Hospital, or using the previous limit.
Then we take the limit
$$\begin{align}\lim_{x\to0}\frac{(1+x)^{1/x}-e}{x}&=\lim_{x\to0}\frac{e^{\frac{\ln(1+x)}{x}}-e}{x}\\&=e\lim_{x\to0}\left[\color{red}{\frac{e^{\frac{\ln(1+x)}{x}-1}-1}{\frac{\ln(1+x)}{x}-1}}\frac{\frac{\ln(1+x)}{x}-1}{x}\right]\\&=e\lim_{x\to0}\frac{\ln(1+x)-x}{x^2}\end{align}$$
The red part went away because it tends to $1$, due to the first two limits above.
Use L'Hospital with the last limit.
$$\lim_{x\to0}\frac{\ln(1+x)-x}{x^2}=\lim_{x\to0}\frac{\frac{1}{1+x}-1}{2x}=\lim_{x\to0}\frac{-1}{2(1+x)}=-\frac{1}{2}$$
Pp..
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